Question 2
Question: The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV, respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, then photo-electric effect is possible for the case of
Options:
A. Both Cs and Li
B. Neither Cs nor Li
C. Li only
D. Cs only
Correct Answer: D
Year: JEE Main 2025 (Online) 22nd January Morning Shift
Solution (as Given in the Source): For photo-electric effect, energy of the photon must be greater than the work function of the metal. We know, $E = \frac{1240}{\lambda} = \frac{1240}{550} \Rightarrow E = 2.25 \text{ eV}$. So, $C_s$ only.,
Step Solution:
1. Identify the condition for the photoelectric effect: Photon Energy ($E$) > Work Function ($\phi$).
2. Use the shortcut formula for photon energy: $E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}$.
3. Substitute the given wavelength: $E = \frac{1240}{550} = 2.254 \text{ eV}$.
4. Compare with Cs: $2.254 \text{ eV} > 1.9 \text{ eV}$ (Possible).,
5. Compare with Li: $2.254 \text{ eV} < 2.5 \text{ eV}$ (Not possible).,
Difficulty Level: Easy
Concept Name: Photoelectric Threshold Energy
Short cut solution: Calculate $E \approx 2.25 \text{ eV}$ using $\frac{1240}{550}$. Since $1.9 < 2.25 < 2.5$, only the metal with the lower work function (Cs) will emit electrons.
Question 3
Question: A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV)
Options:
A. 5 eV
B. 3 eV
C. 2 eV
D. 6 eV
Correct Answer: A
Year: JEE Main 2025 (Online) 22nd January Evening Shift
Solution (as Given in the Source): The photoelectric equation is $K_{max} = \frac{hc}{\lambda} - \phi$. Initially: $2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV} \Rightarrow \frac{hc}{\lambda} = 3 \text{ eV}$. New photon energy with $\frac{\lambda}{2}$ is $\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}$. New $K_{max} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}$.,,
Step Solution:
1. State Einstein’s equation: $K_{max} = \frac{hc}{\lambda} - \phi$.
2. Substitute initial values: $2 = \frac{hc}{\lambda} - 1$.
3. Solve for initial photon energy: $\frac{hc}{\lambda} = 3 \text{ eV}$.
4. Calculate new photon energy for wavelength $\frac{\lambda}{2}$: $E' = 2 \times \left(\frac{hc}{\lambda}\right) = 2 \times 3 = 6 \text{ eV}$.
5. Calculate new maximum kinetic energy: $K'_{max} = 6 - 1 = 5 \text{ eV}$.
Difficulty Level: Medium
Concept Name: Einstein’s Photoelectric Equation
Short cut solution: New $K_{max} = n(K_{old} + \phi) - \phi$, where $n$ is the factor wavelength is reduced by. Here, $n=2$, so $2(2+1) - 1 = 5 \text{ eV}$.,
Question 16
Question: The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is
Options:
A. Red
B. Green
C. Blue
D. Yellow
Correct Answer: C
Year: JEE Main 2025 (Online) 3rd April Morning Shift
Solution (as Given in the Source): For photoemission to occur, $\frac{hc}{\lambda} > \phi$, which implies $\lambda < \frac{hc}{\phi}$. Using $hc \approx 1242 \text{ eV}\cdot\text{nm}$ and $\phi = 3 \text{ eV}$, $\lambda < \frac{1242}{3} \approx 414 \text{ nm}$. Blue light has a wavelength in this range.,,
Step Solution:
1. Set the condition for emission: Photon Energy > Work Function.
2. Express wavelength: $\lambda < \frac{hc}{\phi}$.
3. Calculate threshold wavelength: $\lambda < \frac{1242}{3} = 414 \text{ nm}$.
4. Evaluate visible spectrum: Red ($\approx 700$ nm), Yellow ($\approx 580$ nm), Green ($\approx 530$ nm), Blue ($\approx 400\text{--}450$ nm). (Note: Spectrum ranges are general physics knowledge often used in these problems).
5. Identify the match: Only Blue light satisfies $\lambda < 414 \text{ nm}$.
Difficulty Level: Easy
Concept Name: Threshold Wavelength,
Short cut solution: Threshold $\lambda \approx 1240/3 \approx 413 \text{ nm}$. Since Blue is the only color listed with a wavelength near/below this threshold, it is the only viable option.,
Question 22
Question: The threshold frequency of a metal with work function 6.63 eV is :
Options:
A. $16 \times 10^{15} \text{ Hz}$
B. $16 \times 10^{12} \text{ Hz}$
C. $1.6 \times 10^{12} \text{ Hz}$
D. $1.6 \times 10^{15} \text{ Hz}$
Correct Answer: D
Year: 27-Jan-2024 Shift 2
Solution (as Given in the Source): $\phi_0 = h\nu_0$; $6.63 \times 1.6 \times 10^{-19} = 6.63 \times 10^{-34} \nu_0$; $\nu_0 = \frac{1.6 \times 10^{-19}}{10^{-34}} \Rightarrow \nu_0 = 1.6 \times 10^{15} \text{ Hz}$.
Step Solution:
1. Recall the formula relating work function and threshold frequency: $\phi = h\nu_0$.
2. Convert the work function from eV to Joules by multiplying by $1.6 \times 10^{-19}$: $\phi = 6.63 \times 1.6 \times 10^{-19} \text{ J}$.
3. Equate the energy in Joules to the product of Planck's constant and frequency: $6.63 \times 1.6 \times 10^{-19} = 6.63 \times 10^{-34} \times \nu_0$.
4. Cancel the term $6.63$ from both sides: $1.6 \times 10^{-19} = 10^{-34} \times \nu_0$.
5. Solve for $\nu_0$ by dividing the powers of ten: $\nu_0 = 1.6 \times 10^{15} \text{ Hz}$.
The difficulty level: Easy
The Concept Name: Threshold Frequency Formula
Short cut solution: Since the numerical value of the work function (6.63) matches the numerical value of Planck's constant ($h \approx 6.63 \times 10^{-34}$), they cancel out. Simply divide the conversion factor ($1.6 \times 10^{-19}$) by $10^{-34}$ to get $1.6 \times 10^{15} \text{ Hz}$.
Question 24
Question: The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
Options:
A. 215 nm
B. 414 nm
C. 400 nm
D. 200 nm
Correct Answer: B
Year: 30-Jan-2024 Shift 1
Solution (as Given in the Source): For P.E.E. : $\lambda \leq \frac{1240 \text{ nm-eV}}{3 \text{ eV}}$; $\lambda \leq 413.33 \text{ nm}$; $\lambda_{max} \approx 414 \text{ nm}$ for P.E.E.
Step Solution:
1. Determine the condition for the longest wavelength (threshold wavelength): $\lambda_{max} = \frac{hc}{\phi}$.
2. Use the standard approximation for $hc$ in electron-volt nanometers: $hc \approx 1240 \text{ eV}\cdot\text{nm}$.
3. Substitute the given work function $\phi = 3.0 \text{ eV}$ into the equation: $\lambda_{max} = \frac{1240}{3.0}$.
4. Calculate the result of the division: $\lambda_{max} = 413.33 \text{ nm}$.
5. Round to the nearest integer option: $\lambda_{max} \approx 414 \text{ nm}$.
The difficulty level: Easy
The Concept Name: Threshold Wavelength Calculation
Short cut solution: Use the shortcut formula $\lambda \text{ (nm)} \approx 1240 / \text{Work Function (eV)}$. Dividing 1240 by 3 immediately gives 413.33 nm, which leads to option B.
Question 34
Question: The threshold wavelength for photoelectric emission from a material is 5500Å. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a:
A. 75W infra-red lamp
B. 10W infra-red lamp
C. 75W ultra-violet lamp
D. 10W ultra-violet lamp
Choose the correct answer from the options given below:
Options:
A. B and C only
B. A and D only
C. C only
D. C and D only
Correct Answer: D
Year: 29-Jan-2023 Shift 1
Solution (as Given in the Source): $\lambda < 5500\text{\AA}$ for photoelectric emission. $\lambda_{uv} < 5500\text{\AA}$.
Step Solution:
1. State the primary condition for photoelectric emission: $\lambda_{incident} < \lambda_{threshold}$.
2. Identify the given threshold wavelength: $\lambda_0 = 5500\text{\AA}$.
3. Classify the light sources by wavelength region: Ultraviolet (UV) light has a shorter wavelength ($< 4000\text{\AA}$) and Infrared (IR) light has a longer wavelength ($> 7000\text{\AA}$). (Information generally applied to solve).
4. Compare the regions: Only the UV lamps satisfy the condition because $\lambda_{uv} < 5500\text{\AA}$.
5. Note that emission is independent of power (10W vs 75W) once the frequency/wavelength condition is met: Both C and D will cause emission.
The difficulty level: Medium
The Concept Name: Photoelectric Condition (Wavelength Threshold)
Short cut solution: Emission depends on quality (wavelength), not quantity (power). Since IR wavelength > $5500\text{\AA}$ and UV wavelength < $5500\text{\AA}$, only UV lamps work. Both UV lamp options (C and D) must be part of the answer.
Question 40
Question: If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.86 eV and 2.2 eV. Then choose the correct option
Options:
A. Metal B will not emit photo-electrons
B. Both metals A and B will emit photo-electrons
C. Both metals A and B will not emit photoelectrons
D. Metal A will not emit photo-electrons
Correct Answer: D
Year: JEE Main 31-Jan-2023 Shift 2
Solution (as Given in the Source): $\displaystyle \boldsymbol { \rho } = \mathrm { \small { ~ \frac { h c } { \lambda } = ~ \frac { 1 2 4 0 } { 3 5 0 } e V } } = 3 . 5 4 \mathrm { e V }$. ∴ Only metal B will emit photoelectron.
Step Solution:
1. State the condition for emission: Photon Energy ($E$) must be greater than the Work Function ($\phi$).
2. Use the energy-wavelength shortcut: $E(\text{eV}) = 1240 / \lambda(\text{nm})$.
3. Calculate incident energy: $E = 1240 / 350 = \mathbf{3.54 \text{ eV}}$.
4. Compare with Metal A: $3.54 \text{ eV} < 4.86 \text{ eV}$ (No emission).
5. Compare with Metal B: $3.54 \text{ eV} > 2.2 \text{ eV}$ (Emission occurs).
The difficulty level: Easy
The Concept Name: Photoelectric Threshold Condition
Short cut solution: Calculate $E \approx 3.5 \text{ eV}$. Since $2.2 < 3.5 < 4.86$, only the metal with the lower work function (B) emits electrons, meaning A does not.
Question 42
Question: The threshold frequency of metal is $f_0$. When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5f_0$. the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is:
Options:
A. $\frac{v_1}{v_2} = \frac{1}{2}$
B. $\frac{v_1}{v_2} = \frac{1}{8}$
C. $\frac{v_1}{v_2} = \frac{1}{16}$
D. $\frac{v_1}{v_2} = \frac{1}{4}$
Correct Answer: A
Year: JEE Main 1-Feb-2023 Shift 2
Solution (as Given in the Source): $K_{max} = hf - hf_0$. For $f = 2f_0$: $\frac{1}{2}mv_1^2 = 2hf_0 - hf_0 = hf_0$. For $f = 5f_0$: $\frac{1}{2}mv_2^2 = 5hf_0 - hf_0 = 4hf_0$. $\frac{v_1}{v_2} = \frac{1}{2}$.
Step Solution:
1. Apply Einstein's equation: $\frac{1}{2}mv^2 = h(f - f_0)$.
2. Set up Case 1 ($f = 2f_0$): $\frac{1}{2}mv_1^2 = h(2f_0 - f_0) = \mathbf{hf_0}$.
3. Set up Case 2 ($f = 5f_0$): $\frac{1}{2}mv_2^2 = h(5f_0 - f_0) = \mathbf{4hf_0}$.
4. Divide the two equations: $(\frac{v_1}{v_2})^2 = \frac{hf_0}{4hf_0} = \frac{1}{4}$.
5. Take the square root: $\frac{v_1}{v_2} = \frac{1}{2}$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Velocity $v \propto \sqrt{f - f_{threshold}}$. Ratio = $\sqrt{\frac{2f_0 - f_0}{5f_0 - f_0}} = \sqrt{\frac{1}{4}} = \mathbf{1:2}$.
Question 46
Question: In photoelectric effect:
A. The photocurrent is proportional to the intensity of the incident radiation.
B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.
C. Max K.E with which photoelectrons are emitted depends on the frequency of incident light.
D. The emission of photoelectrons require a minimum threshold intensity of incident radiation.
E. Max. K.E of the photoelectrons is independent of the frequency of the incident light.
Choose the correct answer from the options given below:
Options:
A. B and C only
B. A and C only
C. A and E only
D. A and B only
Correct Answer: B
Year: JEE Main 8-Apr-2023 shift 2
Solution (as Given in the Source): $hv = \phi + (KE)_{max} \Rightarrow (KE)_{max} = hv - \phi$.
Step Solution:
1. Evaluate Statement A: Photocurrent is directly proportional to intensity (Correct).
2. Evaluate Statement B: $K_{max}$ depends on frequency, not intensity (Incorrect).
3. Evaluate Statement C: $K_{max} = h\nu - \phi$, so it is frequency-dependent (Correct).
4. Evaluate Statement D: Threshold depends on frequency ($\nu_0$), not intensity (Incorrect).
5. Evaluate Statement E: $K_{max}$ is explicitly dependent on frequency (Incorrect).
The difficulty level: Easy
The Concept Name: Laws of Photoelectric Emission
Short cut solution: Remember that Intensity affects Quantity (current) and Frequency affects Quality (energy). Only A and C correctly reflect these independent relationships.
Question 52
Question: The difference between threshold wavelengths for two metal surfaces A and B having work function $\Phi_{\text{A}} = 9 \text{ eV}$ and $\Psi_{\text{B}} = 4.5 \text{ eV}$ in nm is: {Given, $hc = 1242 \text{ eV nm}$}
Options:
A. 276
B. 264
C. 540
D. 138
Correct Answer: D
Year: 13-Apr-2023 shift 1
Solution (as Given in the Source): $\Phi = \frac{hc}{\lambda} \Rightarrow \lambda_{\text{A}} = \frac{1242}{9} = 138 \text{ nm}; \lambda_{\text{B}} = \frac{1242}{4.5} = 276 \text{ nm}; \lambda_{\text{B}} - \lambda_{\text{A}} = 276 - 138 = 138 \text{ nm}$.
Step Solution:
1. Recall the formula for threshold wavelength: $\lambda_0 = hc / \Phi$.
2. Calculate $\lambda_{\text{A}}$ for metal A: $\lambda_{\text{A}} = 1242 / 9 = 138 \text{ nm}$.
3. Calculate $\lambda_{\text{B}}$ for metal B: $\lambda_{\text{B}} = 1242 / 4.5 = 276 \text{ nm}$.
4. Find the difference between the two wavelengths: $\Delta\lambda = \lambda_{\text{B}} - \lambda_{\text{A}}$.
5. Perform the subtraction: $276 - 138 = 138 \text{ nm}$.
The difficulty level: Easy
The Concept Name: Threshold Wavelength Calculation
Short cut solution: Since the work function of B is half that of A ($\Phi_{\text{B}} = \Phi_{\text{A}}/2$), its threshold wavelength must be double that of A ($\lambda_{\text{B}} = 2\lambda_{\text{A}}$). The difference $\lambda_{\text{B}} - \lambda_{\text{A}}$ is therefore just $\lambda_{\text{A}}$, which is $1242/9 = 138 \text{ nm}$.
Question 54
Question: When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $v_1$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $v_2$. If $v_2 = xv_1$, the value of x will be
Options: (The source provides this as a numerical value question)
Correct Answer: 2
Year: 24-Jun-2022-Shift-1
Solution (as Given in the Source): Let us say that work function is $\Phi \Rightarrow 2\Phi = \Phi + \frac{1}{2}mv_1^2 \dots (1)$ and $5\Phi = \Phi + \frac{1}{2}mv_2^2 \dots (2)$. From (1) and (2) $\frac{v_2^2}{v_1^2} = \frac{4}{1}$ or $\frac{v_2}{v_1} = 2$.
Step Solution:
1. Use Einstein's equation in terms of work function $\Phi = hf_0$: $hf = \Phi + \frac{1}{2}mv^2$.
2. Set up Case 1 ($f = 2f_0$): $2\Phi = \Phi + \frac{1}{2}mv_1^2 \Rightarrow \frac{1}{2}mv_1^2 = \Phi$.
3. Set up Case 2 ($f = 5f_0$): $5\Phi = \Phi + \frac{1}{2}mv_2^2 \Rightarrow \frac{1}{2}mv_2^2 = 4\Phi$.
4. Divide the kinetic energy equations: $\frac{v_2^2}{v_1^2} = \frac{4\Phi}{\Phi} = 4$.
5. Take the square root to find $x$: $v_2/v_1 = \sqrt{4} = 2$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation (Velocity Ratios)
Short cut solution: Velocity $v \propto \sqrt{f - f_{\text{threshold}}}$. The ratio $x = \sqrt{(5f_0 - f_0) / (2f_0 - f_0)} = \sqrt{4f_0 / f_0} = 2$.
Question 56
Question: The light of two different frequencies whose photons have energies 3.8 eV and 1.46 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :
Options:
A. 1 : 1
B. 2 : 1
C. 4 : 1
D. 1 : 4
Correct Answer: B
Year: 24-Jun-2022-Shift-2
Solution (as Given in the Source): $3.8 = 0.6 + \frac{1}{2}mv_1^2$; $1.4 = 0.6 + \frac{1}{2}mv_2^2 \Rightarrow \frac{v_1^2}{v_2^2} = \frac{3.2}{0.8} = \frac{4}{1} \Rightarrow \frac{v_1}{v_2} = \frac{2}{1}$.
Step Solution:
1. Apply Einstein's equation: $E_{\text{photon}} = \Phi + K_{\text{max}}$.
2. Calculate $K_{\text{max}}$ for Case 1: $K_1 = 3.8 - 0.6 = 3.2 \text{ eV}$.
3. Calculate $K_{\text{max}}$ for Case 2: $K_2 = 1.4 - 0.6 = 0.8 \text{ eV}$ (Note: Solution uses 1.4 eV for calculation based on the question's 1.46 eV typo).
4. Set up the velocity ratio: $(v_1/v_2)^2 = K_1/K_2 = 3.2/0.8$.
5. Solve for the ratio: $(v_1/v_2)^2 = 4 \Rightarrow v_1/v_2 = 2/1$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation (Energy-Velocity relationship)
Short cut solution: Ratio of speeds = $\sqrt{(E_1 - \Phi) / (E_2 - \Phi)} = \sqrt{(3.8 - 0.6) / (1.4 - 0.6)} = \sqrt{3.2 / 0.8} = \sqrt{4} = 2$.
Question 57
Question: Let $K_1$ and $K_2$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_1$ and $\lambda_2$, respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then :
Options:
A. $K_1 > \frac{K_2}{3}$
B. $K_1 < \frac{K_2}{3}$
C. $K_1 = \frac{K}{3}$
D. $K_2 = \frac{K}{3}$
Correct Answer: B
Year: 28-Jun-2022-Shift-2
Solution (as Given in the Source): $K_1 = \frac{hc}{\lambda_1} - \Phi = \frac{hc}{3\lambda_2} - \Phi$ and $K_2 = \frac{hc}{\lambda_2} - \Phi$ from (i) and (ii) we can say $3K_1 = K_2 - 2\Phi \Rightarrow K_1 < \frac{K_2}{3}$
Step Solution:
1. Write the energy equations: $K_1 = \frac{hc}{\lambda_1} - \phi$ and $K_2 = \frac{hc}{\lambda_2} - \phi$.
2. Substitute $\lambda_1 = 3\lambda_2$ into the first equation: $K_1 = \frac{hc}{3\lambda_2} - \phi$.
3. Rearrange to isolate the photon energy term: $K_1 + \phi = \frac{hc}{3\lambda_2}$.
4. Multiply by 3 to compare with $K_2$: $3(K_1 + \phi) = \frac{hc}{\lambda_2}$. Substituting this into the second equation: $3K_1 + 3\phi = K_2 + \phi$.
5. Simplify the relation: $3K_1 = K_2 - 2\phi$. Since $\phi$ is positive, $3K_1$ must be less than $K_2$, meaning $K_1 < \frac{K_2}{3}$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Kinetic energy $K = E - \phi$. If the wavelength triples, the incident energy becomes $1/3$ ($E_1 = E_2/3$). Because the same constant work function ($\phi$) is subtracted from a smaller energy value, the resulting $K_1$ will always be less than $1/3$ of $K_2$.
Question 59
Question: Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.
Reason R: Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements, choose the most appropriate answer from the options given below.
Options:
A. Both A and R are correct and R is the correct explanation of A.
B. Both A and R are correct but R is not the correct explanation of A.
C. A is correct but R is not correct.
D. A is not correct but R is correct.
Correct Answer: B
Year: 29-Jun-2022-Shift-1
Solution (as Given in the Source): To free the electron from metal surface minimum energy required, is equal to the work function of that metal. So Assertion A, is correct. $h\nu = w_0 + K.E._{max}$. if $h\nu = w_0 \Rightarrow K.E._{max} = 0$. Hence reason R, is correct, But R is not the correct explanation of A.
Step Solution:
1. Evaluate Assertion A: By definition, the work function is the minimum energy required to eject an electron; therefore, $E < \phi$ results in no emission. (True).
2. Evaluate Reason R: Use Einstein's equation $K_{max} = E - \phi$.
3. Plug in the condition $E = \phi$: $K_{max} = \phi - \phi = 0$.
4. Verify Statement R: Since $K_{max}$ is 0, the statement is true.
5. Check explanation: While both are facts derived from the same equation, Reason R describes the threshold boundary and does not logically "explain" why emission fails below that boundary in Assertion A. (R is not the explanation).
The difficulty level: Easy
The Concept Name: Photoelectric Threshold Condition
Short cut solution: Both are independent conceptual consequences of the threshold frequency requirement. Since Reason R discusses the equality case and Assertion A discusses the inequality case, they do not have a causal "explanation" relationship.
Question 61
Question: The electric field at a point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15})t + \sin(9 \times 10^{15})t] V m^{-1}$. Given : $h = 4.14 \times 10^{-15} eVs$. If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be
Options:
A. 1.90 eV
B. 3.27 eV
C. 3.60 eV
D. 3.42 eV
Correct Answer: D
Year: 29-Jun-2022-Shift-2
Solution (as Given in the Source): Frequency of EM waves $= \frac{6}{2\pi} \times 10^{15}$ and $\frac{9}{2\pi} \times 10^{15}$. Energy of one photon of these waves $= (4.14 \times 10^{-15} \times \frac{6}{2\pi} \times 10^{15}) eV$ and $(4.14 \times 10^{-15} \times \frac{9}{2\pi} \times 10^{15}) eV = 3.95 eV$ and $5.93 eV \Rightarrow$ Energy of maximum energetic electrons $= 5.93 - 2.50 = 3.43 eV$.
Step Solution:
1. Identify angular frequencies from the $E$ equation: $\omega_1 = 6 \times 10^{15}$ and $\omega_2 = 9 \times 10^{15}$ rad/s.
2. Calculate the higher photon energy using $E = h\nu = h(\frac{\omega}{2\pi})$.
3. Substitute values for $\omega_2$: $E_2 = \frac{4.14 \times 10^{-15} \times 9 \times 10^{15}}{2 \times 3.14}$.
4. Compute photon energy: $E_2 \approx 5.93 \text{ eV}$.
5. Calculate $K_{max} = E_{photon} - \phi$: $5.93 - 2.50 = \mathbf{3.43 \text{ eV}}$ (Matches option D 3.42 eV closest).
The difficulty level: Hard
The Concept Name: Superposition of Waves in Photoelectric Effect
Short cut solution: Max kinetic energy depends only on the highest frequency present in the wave. Ignore the lower $\omega = 6 \times 10^{15}$ and calculate energy for $\omega = 9 \times 10^{15}$. $E \approx 5.93 \text{ eV}$. $K_{max} = 5.93 - 2.5 = 3.43 \text{ eV}$.
Question 63
Question: The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \text{ \AA}$ is 0.42V. If the threshold frequency is $x \times 10^{13} / s$, where $X$ is____ (nearest integer). (Given, speed light $c = 3 \times 10^8 m/s$, Planck's constant $h = 6.63 \times 10^{-34} Js$).
Options: (Numerical value question).
Correct Answer: 35.
Year: 26-Jun-2022-Shift-2.
Solution (as Given in the Source): $\frac{hc}{\lambda} - \phi = KE = eV_{\theta} \Rightarrow \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6630 \times 10^{-10}} - 6.63 \times 10^{-34} f_{th} = 1.6 \times 10^{-19} \times 0.4 \Rightarrow f_{th} \simeq 35.11 \times 10^{13} H$.
Step Solution:
1. Recall Einstein’s Equation: $E_{photon} = \text{Work Function} (\phi) + KE_{max}$, where $\phi = hf_{th}$ and $KE_{max} = eV_0$.
2. Calculate Incident Photon Energy ($E$): $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6630 \times 10^{-10}} = 3 \times 10^{-19} \text{ J}$.
3. Calculate Max Kinetic Energy ($KE$): $KE = 0.42 \times 1.6 \times 10^{-19} = 0.672 \times 10^{-19} \text{ J}$.
4. Find Threshold Energy ($hf_{th}$): $hf_{th} = E - KE = (3 - 0.672) \times 10^{-19} = 2.328 \times 10^{-19} \text{ J}$.
5. Solve for $f_{th}$: $f_{th} = \frac{2.328 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 35.11 \times 10^{13} \text{ Hz}$. Thus, $X \approx 35$.
The difficulty level: Hard
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: In eV, $E \approx \frac{1242}{663} \approx 1.87 \text{ eV}$. Then $\phi \text{ (eV)} = 1.87 - 0.42 = 1.45 \text{ eV}$. Convert to frequency: $f_{th} = \frac{1.45 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 35 \times 10^{13} \text{ Hz}$.
Question 66
Question: A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is : (Take $hc = 1230 eV-nm$).
Options: A. 1.537eV, B. 2.46eV, C. 0.615eV, D. 1.23eV.
Correct Answer: C.
Year: 25-Jul-2022-Shift-1.
Solution (as Given in the Source): $K = \frac{1230}{800} - \Phi$ and $2K = \frac{1230}{500} - \Phi \Rightarrow 2 \times \frac{1230}{800} - 2\Phi = \frac{1230}{500} - \Phi \Rightarrow \Phi = 0.615 \text{ eV}$.
Step Solution:
1. Set up Case 1 Equation: $K = \frac{1230}{800} - \phi = 1.5375 - \phi$.
2. Set up Case 2 Equation: $2K = \frac{1230}{500} - \phi = 2.46 - \phi$.
3. Substitute Case 1 into Case 2: $2(1.5375 - \phi) = 2.46 - \phi$.
4. Solve for $\phi$: $3.075 - 2\phi = 2.46 - \phi$.
5. Final Calculation: $\phi = 3.075 - 2.46 = 0.615 \text{ eV}$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Using the relation $\phi = 2E_1 - E_2$ (when KE doubles): $\phi = 2(\frac{1230}{800}) - \frac{1230}{500} = 3.075 - 2.46 = 0.615 \text{ eV}$.
Question 67
Question: The energy band gap of semiconducting material to produce violet (wavelength $\mu = 4000 \text{ \AA}$) LED is eV . (Round off to the nearest integer)..
Options: (Numerical value question).
Correct Answer: 3.
Year: 25-Jul-2022-Shift-1.
Solution (as Given in the Source): $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}} = \frac{12400}{4000} = 3.1 \text{ eV} \approx 3 \text{ eV}$.
Step Solution:
1. Identify physics principle: Energy band gap ($E_g$) is approximately equal to the energy of the emitted photon ($E = hc/\lambda$).
2. Wavelength Conversion: $\lambda = 4000 \text{ \AA} = 400 \text{ nm}$.
3. Apply Energy Formula: $E = \frac{12400}{\lambda(\text{\AA})} \text{ eV}$.
4. Perform Division: $E = \frac{12400}{4000} = 3.1 \text{ eV}$.
5. Final Approximation: Rounding $3.1 \text{ eV}$ to the nearest integer gives $3$.
The difficulty level: Easy
The Concept Name: Photon Energy and Band Gap relationship
Short cut solution: Use the shortcut $E \approx \frac{1240}{\lambda(\text{nm})}$. For $400 \text{ nm}$, $E = \frac{1240}{400} = 3.1 \text{ eV}$. Rounding gives $3$.
Question 72
Question: Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be
Options:
A. 1 : 2
B. 1 : 3
C. 2 : 3
D. 3 : 2
Correct Answer: C
Year: 28-Jul-2022-Shift-2
Solution (as Given in the Source): $\frac{1}{2} m v_1^2 = 5\phi - \phi$ And, $\frac{1}{2} m v_2^2 = 10\Phi - \Phi \Rightarrow (\frac{\partial \mathbf{v}_1}{\partial \mathbf{v}_2})^2 = \frac{4}{9} \Rightarrow \frac{\partial \mathbf{v}_1}{\partial \mathbf{v}_2} = \frac{2}{3}$
Step Solution:
1. Apply Einstein’s photoelectric equation: $K_{max} = E_{photon} - \phi$.
2. Calculate $K_1$ for Case 1 ($E_1 = 5\phi$): $\frac{1}{2}mv_1^2 = 5\phi - \phi = \mathbf{4\phi}$.
3. Calculate $K_2$ for Case 2 ($E_2 = 10\phi$): $\frac{1}{2}mv_2^2 = 10\phi - \phi = \mathbf{9\phi}$.
4. Set up the ratio of velocities squared: $\frac{v_1^2}{v_2^2} = \frac{4\phi}{9\phi} = \frac{4}{9}$.
5. Take the square root to find the velocity ratio: $\frac{v_1}{v_2} = \frac{2}{3}$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Velocity $v \propto \sqrt{n-1}$ where $n$ is the multiple of the work function. Ratio = $\sqrt{5-1} : \sqrt{10-1} = \sqrt{4} : \sqrt{9} = \mathbf{2:3}$.
Question 73
Question: The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $\lambda.$ To double the kinetic energy, the incident light must have wavelength:
Options:
A. $\frac{hc}{E\lambda - hc}$
B. $\frac{hc\lambda}{E\lambda + hc}$
C. $\frac{h\lambda}{E\lambda + hc}$
D. $\frac{hc\lambda}{E\lambda - hc}$
Correct Answer: B
Year: 29-Jul-2022-Shift-1
Solution (as Given in the Source): $E = \frac{hc}{\lambda} - \phi - (i)$ and $2E = \frac{hc}{\lambda'} - \phi - (ii)$. Subtracting (i) from (ii): $E = hc(\frac{1}{\lambda'} - \frac{1}{\lambda}) \Rightarrow \lambda' = \frac{hc\lambda}{E\lambda + hc}$.
Step Solution:
1. Write Einstein’s equation for the first case: $E = \frac{hc}{\lambda} - \phi$.
2. Write the equation for the second case (doubled kinetic energy): $2E = \frac{hc}{\lambda'} - \phi$.
3. Subtract the first equation from the second to eliminate the work function ($\phi$): $E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}$.
4. Rearrange to isolate $1/\lambda'$: $\frac{1}{\lambda'} = \frac{E}{hc} + \frac{1}{\lambda} = \frac{E\lambda + hc}{hc\lambda}$.
5. Invert the expression to solve for $\lambda'$: $\lambda' = \frac{hc\lambda}{E\lambda + hc}$.
The difficulty level: Medium
The Concept Name: Photoelectric Equation (Wavelength Dependence)
Short cut solution: Use the relation $\frac{1}{\lambda'} = \frac{1}{\lambda} + \frac{\Delta K}{hc}$. Since the increase in kinetic energy $\Delta K$ is $E$, substituting gives $1/\lambda' = \frac{hc + E\lambda}{hc\lambda}$, leading directly to $\lambda' = \frac{hc\lambda}{E\lambda + hc}$.
Question 75
Question: Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is X : y . The value of x is .......
Options: (This was a numerical entry question in the source; the final ratio is $1:3$)
Correct Answer: 1
Year: 26 Feb 2021 Shift 2
Solution (as Given in the Source): $E_{i1} = 2\Phi_0, E_{i2} = 10\Phi_0$. $1/2mv_1^2 = 2\Phi_0 - \Phi_0 = \Phi_0$ and $1/2mv_2^2 = 10\Phi_0 - \Phi_0 = 9\Phi_0$. On dividing, $(v_1/v_2)^2 = 1/9 \Rightarrow v_1/v_2 = 1/3 = x/y \Rightarrow x=1$.
Step Solution:
1. Determine incident energies in terms of work function ($\phi_0$): $E_1 = 2\phi_0$ and $E_2 = 10\phi_0$.
2. Calculate maximum kinetic energy for Case 1: $K_1 = 2\phi_0 - \phi_0 = \mathbf{\phi_0}$.
3. Calculate maximum kinetic energy for Case 2: $K_2 = 10\phi_0 - \phi_0 = \mathbf{9\phi_0}$.
4. Find the ratio of velocities squared: $\frac{v_1^2}{v_2^2} = \frac{\phi_0}{9\phi_0} = \frac{1}{9}$.
5. Find the ratio $v_1/v_2 = 1/3$. Comparing to $x:y$, $x = 1$.
The difficulty level: Easy
The Concept Name: Photoelectric Effect Velocity Ratio
Short cut solution: Velocity ratio = $\sqrt{E_1/\phi - 1} : \sqrt{E_2/\phi - 1} = \sqrt{2-1} : \sqrt{10-1} = \sqrt{1} : \sqrt{9} = \mathbf{1:3}$. Thus, $x=1$.
Question 76
Question: The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491nm is 0.710V . When the incident wavelength is changed to a new value, the stopping potential is 1.43V . The new wavelength is
Options:
A. 309nm
B. 329nm
C. 382nm
D. 400nm
Correct Answer: C
Year: 25 Feb 2021 Shift 2
Solution (as Given in the Source): Given, stopping potential $(V_{\Omega1}) = 0.710 \text{V}$. Incident wavelength $( \lambda_1 ) = 491 \text{nm}$. As, energy $(E) = \frac{1240}{\lambda_1} = \phi_0 + eV$, where, $\phi_0$ is work-function. $\therefore E_1 = \frac{1240}{491} = \phi_0 + 0.71$. And $E_2 = \frac{1240}{\lambda_2} = \phi_0 + 1.43$. Subtracting gives $1240(\frac{1}{\lambda_2} - \frac{1}{491}) = 0.72 \Rightarrow \lambda_2 = 382 \text{nm}$.
Step Solution:
1. Use the energy equation in eV: $E = \frac{1240}{\lambda(\text{nm})}$. For Case 1: $E_1 = \frac{1240}{491} \approx 2.525 \text{ eV}$.
2. Find the work function ($\phi$): $\phi = E_1 - V_1 = 2.525 - 0.710 = 1.815 \text{ eV}$.
3. Apply to Case 2 to find new energy ($E_2$): $E_2 = \phi + V_2 = 1.815 + 1.43 = 3.245 \text{ eV}$.
4. Solve for the new wavelength ($\lambda_2$): $\lambda_2 = \frac{1240}{E_2} = \frac{1240}{3.245}$.
5. Calculate final value: $\lambda_2 \approx 382 \text{ nm}$.
The difficulty level: Medium
The Concept Name: Photoelectric Equation (Energy-Wavelength Relation)
Short cut solution: Use the difference in stopping potentials: $V_2 - V_1 = 1240(\frac{1}{\lambda_2} - \frac{1}{\lambda_1})$. Substituting values: $1.43 - 0.71 = 1240(\frac{1}{\lambda_2} - \frac{1}{491}) \Rightarrow 0.72 = \frac{1240}{\lambda_2} - 2.525$. Thus, $\frac{1240}{\lambda_2} = 3.245 \Rightarrow \lambda_2 \approx 382 \text{ nm}$.
Question 84
Question: Two identical photocathodes receive the light of frequencies $f_1$ and $f_2$ , respectively. If the velocities of the photoelectrons coming out are $v_1$ and $v_2$ respectively, then
Options:
A. $v_1^2 - v_2^2 = \frac{2h}{m}[f_1 - f_2]$
B. $v_1^2 + v_2^2 = \frac{2h}{m}[f_1 + f_2]$
C. $v_1 - v_2 = [\frac{2h}{m}(f_1 + f_2)]^{1/2}$
D. $v_1 - v_2 = [\frac{2h}{m}(f_1 - f_2)]^{1/2}$
Correct Answer: A
Year: 17 Mar 2021 Shift 2
Solution (as Given in the Source): For two identical photocathodes, $\frac{1}{2}mv_1^2 = hf_1 - \Phi \dots (i)$ and $\frac{1}{2}mv_2^2 = hf_2 - \Phi \dots (ii)$. Subtracting Eq. (ii) from Eq. (i) we get $\frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 = hf_1 - hf_2 \Rightarrow v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
Step Solution:
1. State Einstein’s equation for the first frequency: $K_1 = \frac{1}{2}mv_1^2 = hf_1 - \phi$.
2. State the equation for the second frequency: $K_2 = \frac{1}{2}mv_2^2 = hf_2 - \phi$.
3. Subtract the second equation from the first to eliminate the constant work function ($\phi$): $\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$.
4. Rearrange to isolate the velocity difference: $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
The difficulty level: Easy
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Since the work function is identical, the change in maximum kinetic energy is simply the change in photon energy: $\Delta K = h\Delta f$. Replacing $K$ with $\frac{1}{2}mv^2$ yields $\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$ immediately.
Question 99
Question: A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?
Options:
A. 0.96 V
B. 1.25 V
C. 0.24 V
D. 1.5 V
Correct Answer: B
Year: 27 Aug 2021 Shift 2
Solution (as Given in the Source): $eV_1 = \frac{hc}{\lambda_1} - \phi_0$ and $eV_2 = \frac{hc}{\lambda_2} - \phi_0$. So $\frac{hc}{\lambda_1} - eV_1 = \frac{hc}{\lambda_2} - eV_2 \Rightarrow V_2 - V_1 = \frac{hc}{e}(\frac{1}{\lambda_2} - \frac{1}{\lambda_1})$. Substituting values gives $V_2 = 0.48 + 12.4 \times 10^2(2.1 \times 10^{-3} - 1.49 \times 10^{-3}) \approx 1.25 \text{V}$.
Step Solution:
1. Calculate initial photon energy ($E_1$): $E_1 = 1240 / 670.5 \approx 1.849 \text{ eV}$.
2. Determine the work function ($\phi$): $\phi = E_1 - V_1 = 1.849 - 0.48 = 1.369 \text{ eV}$.
3. Calculate the new photon energy ($E_2$): $E_2 = 1240 / 474.6 \approx 2.613 \text{ eV}$.
4. Find the new stopping potential ($V_2$): $V_2 = E_2 - \phi = 2.613 - 1.369$.
5. Final result: $V_2 = 1.244 \text{ V}$, which corresponds to 1.25 V.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Use the stopping potential shift formula: $V_2 = V_1 + 1240(\frac{1}{\lambda_2} - \frac{1}{\lambda_1})$. $V_2 = 0.48 + 1240(\frac{1}{474.6} - \frac{1}{670.5}) \approx 0.48 + (2.61 - 1.85) = 1.24 \text{ V}$.
Question 100
Question: In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work - function $\Phi = 2.5 \text{ eV}$. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. ($\Phi_{\lambda} (\lambda_{h} = 6.63 \times 10^{-34}$ Js., and $c = 3 \times 10^{8} \text{ m s}^{-1}$).
Options: A. 1.3V, B. 1.1V, C. 1.9V, D. 0.6V.
Correct Answer: A.
Year: 26 Aug 2021 Shift 1.
Solution (as Given in the Source): By Einstein’s photoelectric equation, $eV_0 = \frac{hc}{\lambda} - \Phi$. For $\lambda = 280 \text{ nm}$ and $\Phi = 2.5 \text{ eV}$, $e(V_0)_1 = \frac{1240}{280} - 2.5 = 1.93 \text{ eV} \Rightarrow (V_0)_1 = 1.93 \text{ V}$. Similarly, stopping potential for wavelength of 400 nm is $e(V_0)_2 = \frac{1240}{400} - 2.5 \Rightarrow (V_0)_2 = 0.6 \text{ V}$. Change in stopping potential $\Delta V = (V_0)_1 - (V_0)_2 = 1.93 - 0.6 = 1.33 \simeq 1.3 \text{ V}$.
Step Solution:
1. Use the energy-wavelength relation $E = \frac{1240}{\lambda(\text{nm})}$ to find incident energy $E_1$: $1240 / 280 \approx 4.43 \text{ eV}$.
2. Find initial stopping potential $V_1$ using $V_1 = E_1 - \Phi$: $4.43 - 2.5 = 1.93 \text{ V}$.
3. Calculate incident energy $E_2$ for the second wavelength: $1240 / 400 = 3.1 \text{ eV}$.
4. Find new stopping potential $V_2$: $3.1 - 2.5 = 0.6 \text{ V}$.
5. Calculate the difference: $1.93 - 0.6 = 1.33 \text{ V}$, which rounds to 1.3 V.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation.
Short cut solution: $\Delta V = 1240 \times (\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = 1240 \times (\frac{1}{280} - \frac{1}{400}) \approx 4.43 - 3.1 = 1.33 \text{ V}$.
Question 110
Question: When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_{A} \text{ eV}$ and de-Broglie wavelength $\lambda_{A}$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is $T_{B} = (T_{A} - 1.5) \text{ eV}$. If the de-Broglie wavelength of these photoelectrons $\lambda_{B} = 2\lambda_{A}$, then the work function of metal B is:.
Options: A. 4 eV, B. 2 eV, C. 1.5 eV, D. 3 eV.
Correct Answer: A.
Year: 8 Jan 2020 I.
Solution (as Given in the Source): $\lambda = \frac{h}{\sqrt{2m(KE)}} \Rightarrow \lambda \propto \frac{1}{\sqrt{KE}}$. Therefore $\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{T_B}{T_A}} = \sqrt{\frac{T_A - 1.5}{T_A}}$. Given $\frac{\lambda_A}{\lambda_B} = \frac{1}{2}$, solving gives $T_A = 2 \text{ eV}$. Kinetic energy of B is $T_B = 2 - 1.5 = 0.5 \text{ eV}$. Work function of metal B is $\Phi_B = E_B - KE_B = 4.5 - 0.5 = 4 \text{ eV}$.
Step Solution:
1. Relate wavelength to kinetic energy: $\lambda \propto 1/\sqrt{T}$.
2. Set up the ratio: $\lambda_A / \lambda_B = \sqrt{T_B / T_A}$; substituting $\lambda_B = 2\lambda_A$ gives $1/2 = \sqrt{T_B / T_A}$.
3. Square and substitute $T_B = T_A - 1.5$: $1/4 = (T_A - 1.5) / T_A$.
4. Solve for $T_A$: $T_A = 4T_A - 6 \Rightarrow 3T_A = 6 \Rightarrow T_A = 2 \text{ eV}$.
5. Calculate work function of B: $\Phi_B = \text{Photon Energy} - T_B = 4.5 - (2 - 1.5) = \mathbf{4 \text{ eV}}$.
The difficulty level: Hard
The Concept Name: de Broglie Wavelength and Photoelectric Effect.
Short cut solution: Since $\lambda$ doubles, Kinetic Energy must become $1/4$ ($T_B = T_A/4$). Using $T_B = T_A - 1.5$, we get $T_A - 1.5 = 0.25T_A$, so $0.75T_A = 1.5 \Rightarrow T_A = 2 \text{ eV}$. Then $T_B = 0.5 \text{ eV}$ and $\Phi_B = 4.5 - 0.5 = 4 \text{ eV}$.
Question 116
Question: The surface of a metal is illuminated alternately with photons of energies $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is ..
Options: (Numerical Answer Question).
Correct Answer: 2.
Year: Sep. 05, 2020 (II).
Solution (as Given in the Source): $\frac{1}{2}mv_1^2 = 4 - \phi_0$ and $\frac{1}{2}mv_2^2 = 2.5 - \phi_0$. Since $v_1/v_2 = 2$, then $\frac{v_1^2}{v_2^2} = \frac{4 - \phi_0}{2.5 - \phi_0} \Rightarrow 4 = \frac{4 - \phi_0}{2.5 - \phi_0}$. Solving gives $10 - 4\phi_0 = 4 - \phi_0 \Rightarrow \phi_0 = 2 \text{ eV}$.
Step Solution:
1. Apply Einstein's equation for both cases: $K = E - \Phi$.
2. Express the ratio of kinetic energies: $K_1 / K_2 = (v_1 / v_2)^2$.
3. Substitute the given ratio (2): $K_1 / K_2 = 2^2 = 4$.
4. Set up the equation using energies: $(4 - \Phi) / (2.5 - \Phi) = 4$.
5. Solve for $\Phi$: $4(2.5 - \Phi) = 4 - \Phi \Rightarrow 10 - 4\Phi = 4 - \Phi \Rightarrow 3\Phi = 6 \Rightarrow \mathbf{\Phi = 2 \text{ eV}}$.
The difficulty level: Medium
The Concept Name: Photoelectric Equation (Velocity Ratios).
Short cut solution: Use $\Phi = \frac{n^2 E_2 - E_1}{n^2 - 1}$, where $n = 2$. $\Phi = \frac{4(2.5) - 4}{4 - 1} = \frac{10 - 4}{3} = \mathbf{2 \text{ eV}}$.
Question 119
Question: When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:
Options:
A. 0.81 eV
B. 1.02 eV
C. 0.52 eV
D. 0.61 eV
Correct Answer: D
Year: Sep. 03, 2020 (I)
Solution (as Given in the Source): Using equation, $= \frac { h c } { \lambda } - \Phi$. $K E _ { m a x } = \frac { h c } { 5 0 0 } - \Phi \dots ( 1 )$. Again, $3 K E _ { m a x } = \frac { h c } { 2 0 0 } - \Phi \dots ( 2 )$. Dividing equation (2) by (1), $\frac { 3 K E _ { m a x } } { K E _ { m a x } } = \frac { 3 } { 1 } = \frac { \frac { h c } { 2 0 0 } - \phi } { \frac { h c } { 5 0 0 } - \phi }$. Putting the value of $h c = 1 2 3 7 . 5$ and solving we get, work function, $\Phi = 0 . 6 1 \text{ eV}$.
Step Solution:
1. Apply Einstein's equation: $K_{max} = E - \Phi$. Given $K_2 = 3K_1$ for new wavelength.
2. Calculate energy of photons: $E_1 = \frac{1237.5}{500} = 2.475 \text{ eV}$ and $E_2 = \frac{1237.5}{200} = 6.1875 \text{ eV}$.
3. Substitute into the kinetic energy ratio: $\frac{6.1875 - \Phi}{2.475 - \Phi} = 3$.
4. Rearrange and solve for $\Phi$: $6.1875 - \Phi = 7.425 - 3\Phi$.
5. Final Calculation: $2\Phi = 1.2375 \Rightarrow \mathbf{\Phi \approx 0.62 \text{ eV}}$ (Matches 0.61 eV choice).
Difficulty Level: Medium
Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Use the relation $\Phi = \frac{nE_1 - E_2}{n - 1}$ where $n$ is the kinetic energy multiplier. $\Phi = \frac{3(2.48) - 6.19}{2} \approx 0.62 \text{ eV}$.
Question 128
Question: A metal plate of area $1 \times 10^{-4} \text{ m}^2$ is illuminated by a radiation of intensity $16 \text{ mW/m}^2$. The work function of the metal is $5 \text{ eV}$. The energy of the incident photons is $10 \text{ eV}$ and only $10 \%$ of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$]
Options:
A. $10^{14}$ and $10 \text{ eV}$
B. $10^{12}$ and $5 \text{ eV}$
C. $10^{11}$ and $5 \text{ eV}$
D. $10^{10}$ and $5 \text{ eV}$
Correct Answer: C
Year: 10 Jan. 2019 II
Solution (as Given in the Source): using, intensity I = nE. $n =$ no. of photoelectrons. $\Rightarrow 16 \times 10^{-3} = (\frac{n}{t}) \times \frac{10 \times 1.6 \times 10^{-19}}{10^{-4}} : or, \frac{n}{t} = 10^{12}$. So, effective number of photoelectrons ejected per unit time $= 10^{12} \times 10 / 100 = 10^{11}$.
Step Solution:
1. Calculate total power incident on the area: $P = Intensity \times Area = 16 \times 10^{-3} \times 10^{-4} = 1.6 \times 10^{-6} \text{ W}$.
2. Find number of photons arriving per second: $N_{photons} = \frac{P}{E_{photon}} = \frac{1.6 \times 10^{-6}}{10 \times 1.6 \times 10^{-19}} = 10^{12} \text{ s}^{-1}$.
3. Calculate emitted photoelectrons using efficiency ($10\%$): $n_e = 0.1 \times 10^{12} = 10^{11} \text{ s}^{-1}$.
4. Determine maximum kinetic energy: $K_{max} = E_{photon} - \Phi = 10 \text{ eV} - 5 \text{ eV} = 5 \text{ eV}$.
5. Combine results: $10^{11} \text{ electrons/s}$ and $5 \text{ eV}$.
Difficulty Level: Medium
Concept Name: Photon Intensity and Photoelectric Effect
Short cut solution: Rate $= \frac{I \cdot A \cdot \eta}{E_{ph}} = \frac{16 \cdot 10^{-7} \cdot 0.1}{10 \cdot 1.6 \cdot 10^{-19}} = 10^{11}$. Energy is simply the difference $10 - 5 = 5 \text{ eV}$.
Question 129
Question: Surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350$ nm and then, by light of wavelength $\lambda_2 = 540$ nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to: (Energy of photon $\Lambda = \frac{1240}{\lambda(\text{in nm})} \text{ eV}$)
Options:
A. 1.8
B. 2.5
C. 5.6
D. 1.4
Correct Answer: A
Year: 9 Jan. 2019 I
Solution (as Given in the Source): From Einstein's photoelectric equation, ... $\Rightarrow \frac{h c}{\lambda_1} - \frac{\phi}{h c \lambda_2 - \phi} = 4 \Rightarrow \frac{h c}{\lambda_1} - \phi = \frac{4 h c}{\lambda_2} - 4 \phi \Rightarrow \frac{4 h c}{\lambda_2} - \frac{h c}{\lambda_1} = 3 \phi \Rightarrow \Phi = \frac{1}{3} h c \left( \frac{4}{\lambda_2} - \frac{1}{\lambda_1} \right) = \frac{1}{3} \times 1240 \left( \frac{4 \times 350 - 540}{350 \times 540} \right) = 1.8 \text{ eV}$.
Step Solution:
1. Relate kinetic energy to velocity: since speed differs by factor 2, $K_1 = 4K_2$ (because $K = \frac{1}{2}mv^2$).
2. Write Einstein's equations for both cases: $E_1 - \Phi = K_1$ and $E_2 - \Phi = K_2$.
3. Set up the ratio equation: $E_1 - \Phi = 4(E_2 - \Phi)$.
4. Calculate photon energies: $E_1 = 1240/350 = 3.54 \text{ eV}$ and $E_2 = 1240/540 = 2.30 \text{ eV}$.
5. Solve for $\Phi$: $3.54 - \Phi = 4(2.30) - 4\Phi \Rightarrow 3\Phi = 9.20 - 3.54 = 5.66 \Rightarrow \mathbf{\Phi = 1.88 \text{ eV}}$ (Close to 1.8 eV).
Difficulty Level: Hard
Concept Name: Einstein’s Photoelectric Equation (Velocity Ratio)
Short cut solution: Use $\Phi = \frac{n^2 E_2 - E_1}{n^2 - 1}$ where $n=2$. $\Phi = \frac{4(2.30) - 3.54}{3} = 1.88 \text{ eV}$.
Question 130
Question: The magnetic field associated with a light wave is given at the origin by $\tilde{\mathbf{B}} = \mathbf{B}_0 [\sin(3.14 \times 10^7) ct + \sin(6.28 \times 10^7) ct]$. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photoelectrons? ($c = 3 \times 10^8 \text{ m s}^{-1}, h = 6.6 \times 10^{-34} \text{ J-s}$)
Options:
A. 6.82 eV
B. 12.5 eV
C. 8.52 eV
D. 7.72 eV
Correct Answer: D
Year: Not explicitly stated in the source passage, but part of the 2019–2021 collection.
Solution (as Given in the Source): According to question, there are two EM waves with different frequency... To get maximum kinetic energy we take the photon with higher frequency... $v_2 = 10^7 c$. Energy of photon $E_{ph} = \frac{6.6 \times 3 \times 10^{-19}}{1.6 \times 10^{-19}} = 12.375 \text{ eV}$. $K_{max} = 12.375 - 4.7 = 7.675 \text{ eV} \approx 7.7 \text{ eV}$.
Step Solution:
1. Identify the angular frequencies ($\omega$): From the wave equation, $\omega_1 = 3.14 \times 10^7 c$ and $\omega_2 = 6.28 \times 10^7 c$ rad/s.
2. Select the higher frequency for $K_{max}$: Max kinetic energy depends on the highest frequency component, so we use $\omega_2 = (2\pi \times 10^7) \times c$.
3. Calculate the photon energy ($E$): $E = h\nu = h(\frac{\omega}{2\pi}) = \frac{6.6 \times 10^{-34} \times 10^7 \times 3 \times 10^8}{1} = 1.98 \times 10^{-18} \text{ J}$.
4. Convert to eV: $E (\text{eV}) = \frac{1.98 \times 10^{-18}}{1.6 \times 10^{-19}} = 12.375 \text{ eV}$.
5. Calculate $K_{max}$: $K_{max} = E - \phi = 12.375 - 4.7 = 7.675 \text{ eV}$ (rounding to the nearest option D, 7.72 eV).
The difficulty level: Hard
The Concept Name: Photoelectric Equation with Superposed Waves
Short cut solution: Recognize that $\omega = 2\pi \times 10^7 c$, so frequency $f = 10^7 c$. Using $E = hf$, we get $E \approx 12.4 \text{ eV}$. Subtracting the work function: $12.4 - 4.7 = \mathbf{7.7 \text{ eV}}$.
Question 134
Question: In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Options:
A. 1.5 eV
B. 3.0 eV
C. 4.5 eV
D. 15.1 eV
Correct Answer: A
Year: 10 Apr. 2019 I
Solution (as Given in the Source): $K_{max} = E - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = 1237 \left[\frac{1}{260} - \frac{1}{380}\right] = \frac{1237 \times 120}{380 \times 260} = 1.5 \text{ eV}$.
Step Solution:
1. Identify the given values: Incident wavelength $\lambda = 260 \text{ nm}$ and threshold wavelength $\lambda_0 = 380 \text{ nm}$.
2. State the formula: Maximum kinetic energy $K_{max} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.
3. Substitute values using $hc \approx 1237 \text{ eV}\cdot\text{nm}$: $K_{max} = 1237 \times \left( \frac{1}{260} - \frac{1}{380} \right)$.
4. Simplify the fraction: $K_{max} = 1237 \times \left( \frac{380 - 260}{380 \times 260} \right) = 1237 \times \frac{120}{98800}$.
5. Perform final calculation: $K_{max} \approx \mathbf{1.5 \text{ eV}}$.
The difficulty level: Easy
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Use the energy shortcut $E \approx \frac{1240}{\lambda}$. $E_{incident} \approx \frac{1240}{260} \approx 4.77 \text{ eV}$ and $\phi \approx \frac{1240}{380} \approx 3.26 \text{ eV}$. The difference is $4.77 - 3.26 = \mathbf{1.51 \text{ eV}}$.
Question 136
Question: The electric field of light wave is given as $\vec{E} = 10^3 \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat{i} \frac{N}{C}$. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is:
Options:
A. 2.0 V
B. 0.72 V
C. 0.48 V
D. 2.48 V
Correct Answer: C
Year: 9 April 2019 I
Solution (as Given in the Source): Here $w = 2\pi \times 6 \times 10^{14}$ or $f = 6 \times 10^{14} \text{ Hz}$. Wavelength $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{6 \times 10^{14}} = 5000 \text{ \AA}$. Now $E = \frac{12374}{5000} = 2.48 \text{ eV}$. Using $E = w + eV_s \Rightarrow 2.48 = 2 + eV_s$ or $V_s = 0.48 \text{ V}$.
Step Solution:
1. Extract frequency from the equation: The angular frequency $\omega$ is the coefficient of $t$, so $2\pi f = 2\pi \times 6 \times 10^{14}$, giving $f = 6 \times 10^{14} \text{ Hz}$.
2. Calculate wavelength ($\lambda$): $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{6 \times 10^{14}} = 5 \times 10^{-7} \text{ m} = 5000 \text{ \AA}$.
3. Calculate photon energy ($E$): Using the provided conversion $E = \frac{12375}{5000} = 2.475 \text{ eV}$.
4. Apply Einstein's equation for stopping potential: $eV_s = E_{photon} - \phi$.
5. Solve for $V_s$: $V_s = (2.475 - 2) \text{ V} = \mathbf{0.475 \text{ V}} \approx \mathbf{0.48 \text{ V}}$.
The difficulty level: Medium
The Concept Name: Stopping Potential from Wave Equations
Short cut solution: Identify wavelength directly from the wavenumber $k = \frac{2\pi}{\lambda}$. Since $k = \frac{2\pi}{5 \times 10^{-7}}$, then $\lambda = 5 \times 10^{-7} \text{ m} = 500 \text{ nm}$. $E = \frac{1240}{500} = 2.48 \text{ eV}$. Subtracting work function $2 \text{ eV}$ gives $V_s = \mathbf{0.48 \text{ V}}$ directly.
Question 142
Question: The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$, the maximum velocity of the ejected photoelectrons will be:
Options:
A. less than $\sqrt{3}v$
B. $v$
C. more than $\sqrt{3}v$
D. equal to $\sqrt{3}v$
Correct Answer: C
Year: Online April 8, 2017
Solution (as Given in the Source): As the metal surface is same, work function ($\phi$) is same for both the case. Initially $K E_{max} = n h - \phi$ .......(i) After increase $K E_{max} = 3nh - \Phi$ ..... (ii) For work function $\phi^-$ not to be - ve or zero, $v' > \sqrt{3}v$.
Step Solution:
1. Write Einstein's equation for Case 1: $\frac{1}{2}mv^2 = hn - \phi$.
2. Express the initial photon energy: $hn = \frac{1}{2}mv^2 + \phi$.
3. Write the equation for Case 2 (frequency $3n$): $\frac{1}{2}m(v')^2 = 3hn - \phi$.
4. Substitute $hn$ from step 2 into step 3: $\frac{1}{2}m(v')^2 = 3(\frac{1}{2}mv^2 + \phi) - \phi = \frac{3}{2}mv^2 + 2\phi$.
5. Compare the results: $(v')^2 = 3v^2 + \frac{4\phi}{m}$. Since $\phi > 0$, $(v')^2 > 3v^2$, meaning $v' > \sqrt{3}v$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: In the equation $K = E - \phi$, if $E$ is tripled, $K$ increases to $3E - \phi$, which can be rewritten as $3(K + \phi) - \phi = 3K + 2\phi$. Because $K' > 3K$ and $K \propto v^2$, the new velocity $v'$ must be greater than $\sqrt{3}v$.
Question 143
Question: Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$, the speed of the fastest emitted electron will be:
Options:
A. $= v (4/3)^{1/2}$
B. $= v (3/4)^{1/2}$
C. $> v (4/3)^{1/2}$
D. $< v (4/3)^{1/2}$
Correct Answer: C
Year: 2016
Solution (as Given in the Source): $\frac{hc}{\lambda} - hv_0 = \frac{1}{2}mv^2$; $\frac{4hc}{3} - hv_0 = \frac{1}{2}mv'^2$; $v' = v \sqrt{\frac{4v - v_0}{v - v_0}}$; $v' > v \sqrt{4/3}$.
Step Solution:
1. Write the energy equation for Case 1: $K_1 = \frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$.
2. Write the equation for Case 2 ($\lambda' = 3\lambda/4$): $K_2 = \frac{1}{2}m(v')^2 = \frac{4hc}{3\lambda} - \phi$.
3. Multiply the first equation by $4/3$ to match the photon energy term: $\frac{4}{3}K_1 = \frac{4hc}{3\lambda} - \frac{4}{3}\phi$.
4. Establish the relationship between $K_2$ and $K_1$: $K_2 = \frac{4}{3}K_1 + \frac{1}{3}\phi$.
5. Analyze the ratio: Since $\phi$ is positive, $K_2 > \frac{4}{3}K_1$. Taking square roots of the velocity relationship gives $v' > v\sqrt{4/3}$.
The difficulty level: Medium
The Concept Name: Photoelectric Effect (Wavelength dependence)
Short cut solution: When wavelength decreases by a factor $k$, incident energy increases by $1/k$. Since work function remains constant, the kinetic energy increases by a factor greater than $1/k$. Here $1/k = 4/3$, so $v^2$ increases by more than $4/3$, hence $v' > v(4/3)^{1/2}$.
Question 144
Question: A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
Options:
A. $hc/2\lambda$
B. $hc/\lambda$
C. $hc/3\lambda$
D. $3hc/\lambda$
Correct Answer: A
Year: Online April 10, 2016
Solution (as Given in the Source): $K E_{\lambda} = \frac{hc}{\lambda} - \Phi$ (i). $K E_{\lambda/2} = \frac{hc}{\lambda/2} - \Phi$ (ii). From question, $K E_{\lambda/2} = 3(K E_{\lambda}) \Rightarrow \frac{2hc}{\lambda} - \phi = 3 (\frac{hc}{\lambda} - \phi) \Rightarrow 2\phi = \frac{hc}{\lambda} \therefore \phi = \frac{hc}{2\lambda}$.
Step Solution:
1. Express initial kinetic energy: $K_1 = \frac{hc}{\lambda} - \phi$.
2. Express second kinetic energy (energy doubles as wavelength is halved): $K_2 = \frac{2hc}{\lambda} - \phi$.
3. Set up the ratio given in the problem: $K_2 = 3K_1 \Rightarrow \frac{2hc}{\lambda} - \phi = 3(\frac{hc}{\lambda} - \phi)$.
4. Simplify the equation: $\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi$.
5. Solve for $\phi$: $2\phi = \frac{3hc}{\lambda} - \frac{2hc}{\lambda} = \frac{hc}{\lambda}$, thus $\phi = \frac{hc}{2\lambda}$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Use the formula $\phi = \frac{nE_1 - E_2}{n-1}$, where $n$ is the multiplier for kinetic energy (3). $\phi = \frac{3(hc/\lambda) - (2hc/\lambda)}{3 - 1} = \frac{hc/\lambda}{2} = \frac{hc}{2\lambda}$.
Question 145
Question: When photons of wavelength $\lambda_1$ are incident on an isolated sphere, the corresponding stopping potential is found to be $V$. When photons of wavelength $\lambda_2$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $\lambda_3$ is used then find the stopping potential for this case :
Options:
A. $\frac{hc}{e} \left[ \frac{1}{\lambda_3} + \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right]$
B. $\frac{hc}{e} \left[ \frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{1}{\lambda_1} \right]$
C. $\frac{hc}{e} \left[ \frac{1}{\lambda_3} - \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right]$
D. (None)
Correct Answer: D
Year: Online April 9, 2016
Solution (as Given in the Source): From Einstein's photoelectric equation, we have $\frac{hc}{\lambda_1} = \frac{hc}{\lambda_0} + eV \dots(i)$, $\frac{hc}{\lambda_2} = \frac{hc}{\lambda_0} + 3eV \dots(ii)$, $\frac{hc}{\lambda_3} = \frac{hc}{\lambda_0} + eV' \dots(iii)$. From equations $(1) \& (2)$, $\frac{3}{2\lambda_1} - \frac{1}{2\lambda_2} = \frac{1}{\lambda_0}$. Substituting this gives $V' = \frac{hc}{e} \left[ \frac{1}{\lambda_3} - \frac{3}{2\lambda_1} + \frac{1}{2\lambda_2} \right]$.
Step Solution:
1. Write the energy equation for the first two cases: $eV = \frac{hc}{\lambda_1} - \phi$ and $3eV = \frac{hc}{\lambda_2} - \phi$.
2. Multiply the first equation by 3: $3eV = \frac{3hc}{\lambda_1} - 3\phi$.
3. Equate the two expressions for $3eV$: $\frac{hc}{\lambda_2} - \phi = \frac{3hc}{\lambda_1} - 3\phi$.
4. Solve for the work function $\phi$: $2\phi = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2} \implies \phi = \frac{hc}{2} \left( \frac{3}{\lambda_1} - \frac{1}{\lambda_2} \right)$.
5. Substitute $\phi$ into the equation for Case 3 ($V'$): $eV' = \frac{hc}{\lambda_3} - \phi = \frac{hc}{\lambda_3} - \frac{hc}{2} \left( \frac{3}{\lambda_1} - \frac{1}{\lambda_2} \right)$. Simplifying gives $V' = \frac{hc}{e} \left[ \frac{1}{\lambda_3} - \frac{3}{2\lambda_1} + \frac{1}{2\lambda_2} \right]$.
The difficulty level: Hard
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Use the property that stopping potential difference $\Delta V$ is proportional to $\Delta (1/\lambda)$. By eliminating the work function through the ratio of stopping potentials (1:3), you can derive that $V_3 = \frac{hc}{e} [\frac{1}{\lambda_3} - \text{Work function terms}]$, where the work function term is $\frac{1}{2} (\frac{3}{\lambda_1} - \frac{1}{\lambda_2})$.
Question 150
Question: A beam of light has two wavelengths of 4972Å and 6216Å with a total intensity of $3.6 \times 10^{-3} \text{ W m}^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of $1 \text{ cm}^2$ of a clean metallic surface of work function $2.3 \text{ eV}$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately:
Options:
A. $6 \times 10^{11}$
B. $9 \times 10^{11}$
C. $11 \times 10^{11}$
D. $15 \times 10^{11}$
Correct Answer: B
Year: Online April 12, 2014
Solution (as Given in the Source): Intensity associated with each wavelength $= 1.8 \times 10^{-3} \text{ W m}^{-2}$. For $\lambda_1 = 4972\text{\AA}$, $E_1 = 2.493 \text{ eV}$. For $\lambda_2 = 6216\text{\AA}$, $E_2 = 1.994 \text{ eV}$. Since $E_2 < 2.3 \text{ eV}$ (work function), $\lambda_2$ will not contribute. Number of electrons liberated in 2 sec $= 0.45 \times 10^{12} \times 2 = 9 \times 10^{11}$.
Step Solution:
1. Calculate individual intensities: Since intensity is equally distributed, $I_1 = I_2 = 3.6 \times 10^{-3} / 2 = 1.8 \times 10^{-3} \text{ W/m}^2$.
2. Determine photon energies ($E = 12400 / \lambda$): $E_1 \approx 2.49 \text{ eV}$ and $E_2 \approx 1.99 \text{ eV}$.
3. Identify valid emission: Only $E_1 (2.49 \text{ eV})$ is greater than the work function ($2.3 \text{ eV}$); thus, only $\lambda_1$ causes emission.
4. Find number of photons arriving per second for $\lambda_1$: $n = \frac{Intensity \times Area}{Energy} = \frac{1.8 \times 10^{-3} \times 10^{-4}}{2.49 \times 1.6 \times 10^{-19}} \approx \mathbf{0.45 \times 10^{12} \text{ s}^{-1}}$.
5. Calculate total electrons in 2 seconds: Total $= n \times 2 = 0.45 \times 10^{12} \times 2 = \mathbf{9 \times 10^{11}}$.
The difficulty level: Hard
The Concept Name: Photoelectric Threshold and Photon Flux
Short cut solution: Use the threshold condition first to ignore 6216Å. Then, calculating $N = \frac{P \cdot t}{E}$ for only 4972Å yields $9 \times 10^{11}$ immediately.
Question 155
Question: A copper ball of radius 1 cm and work function 4.47eV is irradiated with ultraviolet radiation of wavelength 2500 Å. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is :
Options:
A. $5.5 \times 10^{-13} \text{ C}$
B. $7.5 \times 10^{-13} \text{ C}$
C. $4.5 \times 10^{-12} \text{ C}$
D. $2.5 \times 10^{-11} \text{ C}$
Correct Answer: A
Year: Online April 25, 2013
Solution (as Given in the Source): Not provided in the source excerpt.
Step Solution (Outside source - verified calculation):
1. Calculate Incident Photon Energy ($E$): $E = 12400 / 2500 = \mathbf{4.96 \text{ eV}}$.
2. Find the maximum potential ($V$) of the ball: Emission stops when the ball's potential equals the stopping potential: $V_{max} = E - \phi = 4.96 - 4.47 = \mathbf{0.49 \text{ V}}$.
3. Relate potential to charge ($Q$) for a sphere: $V = \frac{kQ}{R}$, where $k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$ and $R = 0.01 \text{ m}$.
4. Substitute values: $0.49 = \frac{9 \times 10^9 \times Q}{0.01}$.
5. Solve for $Q$: $Q = \frac{0.49 \times 0.01}{9 \times 10^9} \approx \mathbf{5.44 \times 10^{-13} \text{ C}}$, which rounds to $5.5 \times 10^{-13} \text{ C}$.
The difficulty level: Hard
The Concept Name: Photoelectric Effect and Electrostatic Potential
Short cut solution: $Q = \frac{R \cdot V_{stop}}{k}$. Since $V_{stop} \approx 0.5 \text{ V}$ and $R/k = 0.01 / 9 \times 10^9 \approx 1.1 \times 10^{-12}$, the charge $Q \approx 0.5 \times 1.1 \times 10^{-12} \approx \mathbf{0.55 \times 10^{-12} \text{ C}}$.
Question 157
Question: Photoelectrons are ejected from a metal when light of frequency $\nu$ falls on it. Pick out the wrong statement from the following.
Options:
A. No electrons are emitted if $\nu$ is less than $W/h$, where $W$ is the work function of the metal.
B. The ejection of the photoelectrons is instantaneous.
C. The maximum energy of the photoelectrons is $h\nu$.
D. The maximum energy of the photoelectrons is independent of the intensity of the light.
Correct Answer: C
Year: Online May 26, 2012
Solution (as Given in the Source): According to photo-electric equation: $K.E_{max} = h\nu - h\nu_0$ (Work function). Some sort of energy is used in ejecting the photoelectrons.
Step Solution:
1. State Einstein's Equation: $K_{max} = h\nu - W$, where $h\nu$ is photon energy and $W$ is the work function.
2. Evaluate Option A: For emission, $h\nu \ge W \implies \nu \ge W/h$. Thus, if $\nu < W/h$, no emission occurs. (Correct statement).
3. Evaluate Option B: Photoelectric effect is scientifically established as an instantaneous process. (Correct statement).
4. Evaluate Option C: From the energy equation, $K_{max}$ is always $h\nu$ minus the work function. Since $W > 0$, $K_{max}$ must be less than $h\nu$. (Wrong statement).
5. Evaluate Option D: Kinetic energy depends on the frequency of light, not the intensity. (Correct statement).
The difficulty level: Easy
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: In the equation $K_{max} = E_{photon} - W$, because a part of the energy is used to overcome the work function, the final kinetic energy can never equal the total incident energy ($h\nu$).
Question 161
Question: The surface of a metal is illuminted with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : ($hc = 1240 \text{ eV}\cdot\text{nm}$)
Options:
A. 1.41 eV
B. 1.51 eV
C. 1.68 eV
D. 3.09 eV
Correct Answer: A
Year: 2009
Solution (as Given in the Source): Wavelength of incident light, $\lambda = 400 \text{ nm}$; $hc = 1240 \text{ eV}\cdot\text{nm}$; $K.E = 1.68 \text{ eV}$. Using Einstein's photoelectric equation $\frac{hc}{\lambda} - W = K.E \Rightarrow W = \frac{hc}{\lambda} - K.E = \frac{1240}{400} - 1.68 = 3.1 - 1.68 = 1.41 \text{ eV}$.
Step Solution:
1. Identify variables: $\lambda = 400 \text{ nm}$, $K_{max} = 1.68 \text{ eV}$, $hc = 1240 \text{ eV}\cdot\text{nm}$.
2. Set up the formula: $W = E_{photon} - K_{max}$.
3. Calculate Photon Energy ($E$): $E = \frac{1240}{400} = 3.1 \text{ eV}$.
4. Substitute values: $W = 3.1 - 1.68$.
5. Final Calculation: $W = 1.42 \text{ eV}$ (rounding as per source/options gives 1.41 eV).
The difficulty level: Easy
The Concept Name: Photoelectric Work Function
Short cut solution: Use the relation $W = \frac{1240}{\lambda} - K_{max}$. Quickly divide 1240 by 400 to get 3.1, then subtract 1.68 to get 1.42 eV.
Question 165
Question: The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in
Options:
A. ultra-violet region
B. infra-red region
C. visible region
D. X-ray region
Correct Answer: A
Year: 2006
Solution (as Given in the Source): Work function, $\Phi = 6.2 \text{ eV} = 6.2 \times 1.6 \times 10^{-19} \text{ J}$. Stopping potential, $V = 5 \text{ volt}$. From the Einstein's photoelectric equation $\frac{hc}{\lambda} - \phi = eV_0 \Rightarrow \lambda = \frac{hc}{\phi + eV_0} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} (6.2 + 5)} \approx 10^{-7} \text{ m}$. This range lies in ultra violet range.
Step Solution:
1. Identify given energies: Threshold energy (Work function $W$) = 6.2 eV; Stopping Potential ($V_s$) = 5 V.
2. Calculate Max Kinetic Energy: $K_{max} = e \times V_s = 5 \text{ eV}$.
3. Determine Incident Photon Energy ($E$): $E = W + K_{max} = 6.2 + 5 = 11.2 \text{ eV}$.
4. Convert Energy to Wavelength ($\lambda$): $\lambda = \frac{1240}{11.2} \approx 110.7 \text{ nm}$.
5. Region Classification: Wavelengths between $\sim 10$ nm and 400 nm are in the ultraviolet region.
The difficulty level: Medium
The Concept Name: Electromagnetic Spectrum Identification
Short cut solution: Sum the work function and stopping potential to find total photon energy ($11.2 \text{ eV}$). Since $3.1 \text{ eV}$ is the limit for visible light ($\sim 400 \text{ nm}$), an energy as high as $11.2 \text{ eV}$ must be in the UV region.
Question 172
Question: The work function of a substance is $4.0 \text{ eV}.$ The longest wavelength of light that can cause photoelectron emission from this substance is approximately
Options:
A. 310 nm
B. 400 nm
C. 540 nm
D. 220 nm
Correct Answer: A
Year: 2004
Solution (as Given in the Source): Work function of metal $(\phi)$ is given by $\phi = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{\phi} \Rightarrow \lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.6 \times 10^{-19}} = 310 \text{ nm}$.
Step Solution:
1. Identify the threshold (longest) wavelength formula: $\lambda = \frac{hc}{\phi}$.
2. Convert the work function from eV to Joules: $4 \text{ eV} \times 1.6 \times 10^{-19} = 6.4 \times 10^{-19} \text{ J}$.
3. Substitute constants ($h \approx 6.63 \times 10^{-34} \text{ Js}, c \approx 3 \times 10^8 \text{ m/s}$): $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.4 \times 10^{-19}}$.
4. Calculate the product and divide: $\lambda = \frac{19.89 \times 10^{-26}}{6.4 \times 10^{-19}} = 3.107 \times 10^{-7} \text{ m}$.
5. Convert the result to nanometers: $3.107 \times 10^{-7} \text{ m} \approx 310 \text{ nm}$.
The difficulty level: Easy
The Concept Name: Threshold Wavelength Calculation
Short cut solution: Use the approximation $\lambda (\text{nm}) \approx \frac{1240}{\phi (\text{eV})}$. Substituting the given value: $\lambda \approx \frac{1240}{4} = 310 \text{ nm}$.
Question 173
Question: Two identical photocathodes receive light of frequencies $f_1$ and $f_2$. If the velocites of the photo electrons (of mass m) coming out are respectively $v_1$ and $v_2$, then
Options:
A. $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$
B. $v_1 + v_2 = [\frac{2h}{m}(f_1 + f_2)]^{1/2}$
C. $v_1^2 + v_2^2 = \frac{2h}{m}(f_1 + f_2)$
D. $v_1 - v_2 = [\frac{2h}{m}(f_1 - f_2)]^{1/2}$
Correct Answer: A
Year: 2003
Solution (as Given in the Source): $hf_1 - W = \frac{1}{2}mv_1^2$ (i) and $hf_2 - W = \frac{1}{2}mv_2^2$ (ii). Subtracting (ii) from (i) we get $h(f_1 - f_2) = \frac{m}{2}(v_1^2 - v_2^2) \therefore v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
Step Solution:
1. Apply Einstein's equation for frequency $f_1$: $\frac{1}{2}mv_1^2 = hf_1 - W$.
2. Apply Einstein's equation for frequency $f_2$: $\frac{1}{2}mv_2^2 = hf_2 - W$.
3. Subtract the second equation from the first to eliminate the constant work function $W$: $\frac{1}{2}m(v_1^2 - v_2^2) = h(f_1 - f_2)$.
4. Multiply both sides by 2: $m(v_1^2 - v_2^2) = 2h(f_1 - f_2)$.
5. Divide by mass $m$ to isolate the velocity terms: $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
The difficulty level: Medium
The Concept Name: Einstein’s Photoelectric Equation
Short cut solution: Since the change in maximum kinetic energy $(\Delta K)$ for a single metal is equal to the change in incident photon energy $(\Delta E)$, simply equate $\frac{1}{2}m\Delta(v^2) = h\Delta f$. This directly yields $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
Question 175
Question: Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of the wavelengths is nearest to
Options:
A. 1 : 2
B. 4 : 1
C. 2 : 1
D. 1 : 4
Correct Answer: C
Year: 2002
Solution (as Given in the Source): $\frac{E_{Na}}{E_{Cu}} = \frac{\lambda_{Cu}}{\lambda_{Na}} \Rightarrow \frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{E_{Cu}}{E_{Na}} = \frac{4.5}{2.3} \approx \frac{2}{1}$.
Step Solution:
1. Relate work function energy ($E$) to threshold wavelength ($\lambda$): $E = \frac{hc}{\lambda}$.
2. Establish the inverse relationship for two materials: $\frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1}$.
3. Set up the ratio for Sodium ($Na$) and Copper ($Cu$): $\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{E_{Cu}}{E_{Na}}$.
4. Substitute the given work function values: $\frac{\lambda_{Na}}{\lambda_{Cu}} = \frac{4.5 \text{ eV}}{2.3 \text{ eV}}$.
5. Perform the division: $4.5 / 2.3 = 1.956 \approx 2$. The ratio is nearest to $2:1$.
The difficulty level: Easy
The Concept Name: Work Function and Threshold Wavelength Relationship
Short cut solution: Since wavelength is inversely proportional to work function, and Copper's work function is nearly double that of Sodium ($4.5 \approx 2 \times 2.3$), Sodium's threshold wavelength must be approximately double that of Copper ($2:1$).