Dual Nature of Radiation

$ e V_s=h v=\phi_0 $

$ \begin{aligned} \Rightarrow \quad V_s & =\frac{h}{e} v-\frac{\phi_0}{e} \\ \therefore \text { Slope } & =\frac{h}{e}=\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19}} \\ & =4.125 \times 10^{-15} \mathrm{Js} \mathrm{C}^{-1} \end{aligned} $

Work Done = Change in Kinetic Energy

The work needed is the same as how much the electron's kinetic energy increases.

At the start, the electron is at rest, so its initial kinetic energy $\mathrm{KE}_i = 0$.

After speeding up, the electron’s final kinetic energy $\mathrm{KE}_f = W$, where $W$ is the work done.

So, $W = \mathrm{KE}_f - \mathrm{KE}_i = \mathrm{KE}_f = W$

Relating de-Broglie Wavelength to Kinetic Energy

The de-Broglie wavelength formula is:

$ \lambda_d = \frac{h}{mv} = \frac{h}{p} $

For an electron (or any particle), $p = \sqrt{2m\mathrm{KE}}$.

So, $ \lambda_d = \frac{h}{\sqrt{2m \mathrm{KE}}} $

From above, $\mathrm{KE} = W$ (work done).

Therefore, $ \lambda_d = \frac{h}{\sqrt{2mW}} $

Now, solve for $W$: $ \lambda_d^2 = \frac{h^2}{2mW} $ $ W = \frac{h^2}{\lambda_d^2 \cdot 2m} $

Plugging in the Numbers

Given:

• Planck's constant, $h = 6.6 \times 10^{-34}\ \mathrm{Js}$
• Mass of electron, $m = 9 \times 10^{-31}\ \mathrm{kg}$
• de-Broglie wavelength, $\lambda_d = 6600\ \text{Å} = 6.6 \times 10^{-7}\ \mathrm{m}$

So, $ W = \frac{(6.6 \times 10^{-34})^2}{(6.6 \times 10^{-7})^2 \times 2 \times 9.0 \times 10^{-31}}\ \mathrm{J} $

Now calculate each step:

Top: $(6.6 \times 10^{-34})^2 = 43.56 \times 10^{-68}$

Bottom: $(6.6 \times 10^{-7})^2 \times 2 \times 9.0 \times 10^{-31} = (43.56 \times 10^{-14}) \times 18.0 \times 10^{-31} = 784.08 \times 10^{-45}$

So, $ W = \frac{43.56 \times 10^{-68}}{784.08 \times 10^{-45}} $

This simplifies to: $ W = \frac{1}{18.0} \times 10^{-23}\ \mathrm{J} $

So, the answer is: $ W = 5.56 \times 10^{-25}\ \mathrm{J} $

The photoelectric effect equation relates the energy of incident photons, the work function of the material, and the maximum kinetic energy of emitted photoelectrons:

$E_{photon} = \phi + KE_{max}$

The maximum kinetic energy ($KE_{max}$) of the emitted photoelectrons is related to their maximum velocity ($v_{max}$) by the formula:

$KE_{max} = \frac{1}{2} m v_{max}^2$

The stopping potential ($V_s$) is the potential difference required to stop the most energetic photoelectrons. The work done by the stopping potential must be equal to the maximum kinetic energy of the electrons:

$KE_{max} = e V_s$

We are given:

• Work function ($\phi$) = 1.5 eV (Note: This is not directly needed to find the stopping potential if $v_{max}$ is given, but it would be needed to find the photon energy.)

• Maximum velocity ($v_{max}$) = $8 \times 10^5 \mathrm{~ms}^{-1}$

• Mass of electron ($m$) = $9 \times 10^{-31} \mathrm{~kg}$

• Charge of electron ($e$) = $1.6 \times 10^{-19} \mathrm{~C}$

First, calculate the maximum kinetic energy ($KE_{max}$) using the given maximum velocity:

$KE_{max} = \frac{1}{2} m v_{max}^2$

$KE_{max} = \frac{1}{2} \times (9 \times 10^{-31} \mathrm{~kg}) \times (8 \times 10^5 \mathrm{~ms}^{-1})^2$

$KE_{max} = \frac{1}{2} \times 9 \times 10^{-31} \times (64 \times 10^{10})$

$KE_{max} = \frac{1}{2} \times 9 \times 64 \times 10^{-31+10}$

$KE_{max} = 9 \times 32 \times 10^{-21}$

$KE_{max} = 288 \times 10^{-21} \mathrm{~J}$

Now, use the relationship between maximum kinetic energy and stopping potential ($KE_{max} = e V_s$) to find the stopping potential ($V_s$):

$V_s = \frac{KE_{max}}{e}$

$V_s = \frac{288 \times 10^{-21} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~C}}$

$V_s = \frac{288}{1.6} \times \frac{10^{-21}}{10^{-19}}$

$V_s = \frac{2880}{16} \times 10^{-2}$

$V_s = 180 \times 10^{-2}$

$V_s = 1.8 \mathrm{~V}$

The stopping potential of the photoelectrons is 1.8 V.

The final answer is $\boxed{\text{1.8 V}}$.

To find the smallest (minimum) wavelength of X-rays made by 20 kV electrons, use this formula:

$ \lambda_{min} = \frac{hc}{E} = \frac{hc}{eV} $ where:
$ h $ = Planck's constant = $ 6.626 \times 10^{-34} $ Js
$ c $ = speed of light = $ 3 \times 10^8 $ m/s
$ e $ = charge of electron = $ 1.6 \times 10^{-19} $ C
$ V $ = voltage = 20,000 V = $ 20 \times 10^3 $ V

Plug in the numbers:

$ \lambda_{min} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 20 \times 10^3} $

Solve the calculation:

$ \lambda_{min} = 0.062 \times 10^{-9}~\mathrm{m} $

We can also write this as:

$ \lambda_{min} = 0.062~\mathrm{nm} $

When light hits a photosensitive material, electrons are released. The energy of the incoming light (photon) is used to release these electrons and provide them with some extra energy.

Einstein’s photoelectric equation helps us understand this process:

$ E_{\max} = h \nu - \phi_0 $

This means: the maximum energy of the emitted electrons equals the energy of the incoming photon minus the work function ($\phi_0$), which is the energy needed to release the electron.

The stopping potential ($V_0$) is linked to this energy by:

$ eV_0 = h \nu - \phi_0 $

or

$ V_0 = \frac{h \nu}{e} - \frac{\phi_0}{e} $

First, use the given values for the first case:

For photon energy 3.1 eV and stopping potential 1.7 V:

$ V_0 = 3.1 - \frac{\phi_0}{e} $

$ 1.7 = 3.1 - \frac{\phi_0}{e} $

To find the work function divided by $e$:

$ \frac{\phi_0}{e} = 3.1 - 1.7 = 1.4 $

So, $ \frac{\phi_0}{e} = 1.4 \text{ volt} $

Now, use this value for the second case:

For photon energy 2.5 eV:

$ V_0 = 2.5 - \frac{\phi_0}{e} $

$ V_0 = 2.5 - 1.4 = 1.1 \text{ V} $

The maximum kinetic energy

$ \begin{aligned} & \mathrm{KE}_{\max }=E-\phi \\ & \mathrm{KE}_{\max }=4.5 \mathrm{eV}-3 \mathrm{eV} \\ & \mathrm{KE}_{\max }=1.5 \mathrm{eV} \end{aligned} $

The de-Broglie wavelength is,

$ \begin{aligned} & \lambda=\frac{12.27}{\sqrt{K E_{\max }}}\mathop {\rm{A}}\limits^{\rm{o}} \\ & \lambda=\frac{12.27}{\sqrt{1.5}} \mathop {\rm{A}}\limits^{\rm{o}} \Rightarrow \lambda \approx \frac{12.27}{1.2247} \mathop {\rm{A}}\limits^{\rm{o}} \\ & \lambda \approx 10.018 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

The de-Broglie wavelength associated with the photoelectrons is approximately $10.02 \mathop {\rm{A}}\limits^{\rm{o}}$.

Given,

$ \phi_{A}: \phi_{B}=2: 3 \Rightarrow \phi_{A}=2 k, \phi_{B}=3 k $

Using Einstein's photoelectric equation

$ e V_{0}=h v-h v_{0} $

where, $ h= $ Planck's constant, $ v= $ frequency of incident light

$ v_{0}= $ threshold frequency, $ V_{0}= $ stopping potential

$ V_{0}=\frac{h}{e} v-\frac{h}{e} v_{0} $

Now comparing Eq. (i) with $ y=m x+C $

Here, slope $ m=\frac{h}{e} $.

The slopes $ x $ and $ y $ of the graphs drawn between the stopping potential and the frequency of incident light for surfaces

$A$ and $B$ depend only on the constant $h$ and $e$ not on the work function of $A$ and $B$. So, $x$ and $y$ are the same for both metals.

$ x=y=\frac{h}{e} $

Hence, $x: y=1: 1$

The wave picture of light fails to explain the photoelectric effect.

The photoelectric effect is the phenomenon where electrons are ejected from a material's surface when it is exposed to light of certain frequencies. According to the classical wave theory of light, increasing the intensity of light (regardless of frequency) should increase the energy of emitted electrons. However, experiments show that there is a minimum threshold frequency below which no electrons are emitted, regardless of light intensity. Moreover, the maximum kinetic energy of the emitted electrons depends on the frequency of the incident light, not its intensity.

Albert Einstein explained the photoelectric effect by proposing that light consists of discrete packets of energy called photons. Each photon has an energy given by the equation:

$ E = hf $

where:

$ E $ is the energy of the photon,

$ h $ is Planck's constant ($6.626 \times 10^{-34} \, \text{Js}$),

$ f $ is the frequency of the light.

Thus, only photons with enough energy (i.e., with frequency above the threshold frequency) can eject electrons, thereby providing a particle-like explanation for light, which is not accounted for by its wave-like nature.

In contrast, the wave picture of light successfully explains phenomena like interference, diffraction, and polarization of light. Interference and diffraction involve the superposition of light waves to produce patterns, and polarization relates to the orientation of light waves, all of which align with the wave nature of light.

Given:

Power of the bulb = $60 \, \text{W}$

Efficiency = $16\%$

Calculation of Actual Power:

The actual power is determined by multiplying the efficiency with the total power:

$ P = \frac{16}{100} \times 60 = 9.6 \, \text{W} $

Intensity of Electromagnetic Radiation:

The intensity, $I$, of the electromagnetic radiation is given by the formula:

$ I = \frac{1}{2} \varepsilon_{0} c E_{0}^{2} $

where:

$c$ is the speed of light.

$E_{0}$ is the peak electric field.

Relation of Intensity with Power and Area:

The intensity is also expressed as power per unit area:

$ I = \frac{\text{Power}}{\text{Area}} = \frac{P}{4 \pi r^{2}} $

Equating the two expressions for intensity:

$ \frac{P}{4 \pi r^{2}} = \frac{1}{2} \varepsilon_{0} c E_{0}^{2} $

Solving for $E_{0}^{2}$:

$ E_{0}^{2} = \frac{2P}{4 \pi \varepsilon_{0} c r^{2}} $

Substituting the Known Values:

$r = 2 \, \text{m}$

$c = 3 \times 10^8 \, \text{m/s}$

$\varepsilon_{0} = \frac{1}{4 \pi \times 9 \times 10^9}$

Substituting these values into the equation for $E_{0}^{2}$:

$ E_{0}^{2} = \frac{2 \times 9.6 \times 9 \times 10^9}{4 \times \pi \times \left(\frac{1}{4 \pi \times 9 \times 10^9}\right) \times (3 \times 10^8) \times 2^2} $

Simplifying further:

$ E_{0}^{2} = \frac{2 \times 9.6 \times 9 \times 10^9}{4 \times 3 \times 10^8} $

$ E_{0}^{2} = 144 $

Hence, the peak value of the electric field is:

$ E_{0} = \sqrt{144} = 12 \, \text{V/m} $

Given,

Work function, $ \phi=1.1 \mathrm{eV} $

Using photoelectric equation for each incident radiation.

$ E=\frac{1}{2} m v^{2}+\phi $

where $ E $ is energy of incident radiation $ \frac{1}{2} m v^{2}= $ kinetic energy of photoelectron

$ \begin{array}{l} E_{1}=\frac{1}{2} m v_{1}^{2}+\phi \\\\ E_{2}=\frac{1}{2} m v_{2}^{2}+\phi \end{array} $

From Eqs. (ii) and (iii)

$ \begin{aligned} \frac{v_{1}^{2}}{v_{2}^{2}} & =\frac{E_{1}-\phi}{E_{2}-\phi} \\\\ & =\frac{1.5-1.1}{2-1.1} \\\\ \frac{v_{1}^{2}}{v_{2}^{2}} & =\frac{0.4}{0.9} \Rightarrow \frac{v_{1}}{v_{2}}=\frac{2}{3} \\\\ v_{1} & : v_{2}=2: 3 \end{aligned} $

Given, de-Broglie wavelength of proton $ =2 \times $ de-Broglie wavelength of $ \alpha $-particle

de-Broglie wavelength in term of kinetic energy $ E $ and mass $ m $ is,

$ \lambda=\frac{h}{\sqrt{2 m E}} $

Now, $ \quad \lambda_{p}=2 \lambda_{\alpha} $

$ \Rightarrow \quad \frac{h}{\sqrt{2 m E_{1}}}=\frac{2 h}{\sqrt{2(4 m) E_{2}}} $

where $ E_{1} $ and $ E_{2} $ are KE respectively

$ \frac{1}{\sqrt{E_{1}}}=\frac{1}{\sqrt{E_{2}}} $

$ \therefore \quad \frac{E_{1}}{E_{2}}=\frac{1}{1} $

Ratio is $ 1: 1 $

Energy of neutron at temperature,

$ T=\frac{3}{2} k_B T $

de-Broglie wavelength

$ =\frac{h}{\sqrt{2 m E}} \propto \frac{h}{\sqrt{2 \times m_n \times T}} $

Here, $T_1=(273+77) \mathrm{K}=350 \mathrm{~K}$

$ \begin{aligned} T_2 & =(273+1127) \mathrm{K}=1400 \mathrm{~K} \\\\ \therefore \frac{\lambda_2}{\lambda_1} & =\frac{h}{\sqrt{2 m_n \times 1400}} \times \frac{\sqrt{2 m_n \times 350}}{h} \\\\ \lambda_2 & =\lambda \times \frac{1}{2} \end{aligned} $

The relationship between the cut-off voltage $ V_0 $ and the frequency $ f $ of incident light in a photoelectric experiment is given by the equation:

$ eV_0 = hf - \phi $

where:

$ e $ is the elementary charge ($1.6 \times 10^{-19} \, \mathrm{C}$),

$ h $ is Planck's constant ($6.63 \times 10^{-34} \, \mathrm{Js}$),

$ \phi $ is the work function of the material.

Rearranging the equation provides:

$ V_0 = \frac{h}{e} f - \frac{\phi}{e} $

The slope $ m $ of the graph of $ V_0 $ versus $ f $ is given by the factor of $ f $, which is $ \frac{h}{e} $.

Substituting the given values:

$ m = \frac{h}{e} = \frac{6.63 \times 10^{-34} \, \mathrm{Js}}{1.6 \times 10^{-19} \, \mathrm{C}} $

Calculating this gives:

$ m = \frac{6.63}{1.6} \times 10^{-15} \approx 4.14375 \times 10^{-15} \, \mathrm{Vs} $

Thus, the correct option is:

Option A: $ 414 \times 10^{-15} \, \mathrm{Vs} $

The similar materials will have same stopping potential and the different materials will have different stopping potential whatever may be the intensity of incident radiation.

So, curve $(1,2)$ and $(3,4)$ represent same materials.

So, option (d) is the correct option.

Given, Radiation's wavelength, $\lambda=400 \mathrm{~nm}$ Work function, $\phi=2.2 \mathrm{eV}$ Energy of incident photon is given by

$ E=\frac{1240}{\lambda(\mathrm{~nm})} \mathrm{eV}=\frac{1240}{400}=3.1 \mathrm{eV} $

Maximum kinetic energy of ejected electron

$ \mathrm{KE}=E-\phi=3.1-2.2=0.9 \mathrm{eV} $

Potential is energy/unit charge

$ \begin{aligned} V & =\frac{0.9 \mathrm{eV}}{e} \\ \Rightarrow \quad V & =0.9 \mathrm{~V} \end{aligned} $

Hint

Energy of $n$-photon,

$ E_n=n h v=n h f $

Energy of $2 n$ photons,

$ E_{2 n}=2 n h(3 f)=6 n h f $

$ \frac{\text { Average radiation intensity on surface } A}{\text { Average radiation intensity on surface } B} $

$ \begin{aligned} & =\frac{E_n / t_1}{E_{2 n} / t_2}=\frac{E_n \times t_2}{E_{2 n} \times t_1} \\ & =\frac{(n h f) \times 4 t}{6 n h f \times t}=\frac{4}{6}=\frac{2}{3} \text { or } 2: 3 \end{aligned} $

Energy of photon released by the transition of an electron from second excited state $\left(n_2=3\right)$ of ground state $\left(n_1=1\right)$ is given as

$ \begin{aligned} \Delta E & =h v=E_3-E_1 \\ & =-\frac{13.6}{3^2}-\left(-\frac{13.6}{1^2}\right) \\ & =13.6-\frac{13.6}{9} \\ & =\frac{8}{9} \times 13.6 \mathrm{eV} \end{aligned} $

According to Einstein's photoelectric equation

$ \begin{aligned} \mathrm{KE} & =h v-\phi_0 \\ & =\frac{8}{9} \times 13.6-314 \approx 10 \mathrm{eV} \end{aligned} $

de-Broglie wavelength of the most energetic electron,

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 m(K E)}} \\ & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.31 \times 10^{-31} \times 10 \times 1.6 \times 10^{-19}}} \\ & =4.09 \times 10^{-10} \mathrm{~m}=4.09 \mathop {\rm{A}}\limits^{\rm{o}}=\simeq 4 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Given speed of moving particle

$ y=0.8 \mathrm{c} $

We know that, de-Broglie wavelength associated with moving particle

$ \begin{aligned} \lambda & =\frac{h}{m v} \\ \lambda & =\frac{h}{m \times 0.8 c} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

de-Broglie wavelength of photon

$ \lambda=\frac{h}{m c} $

Energy of photon, $E=\frac{h c}{\lambda}$

Kinetic energy of particle $E^{\prime}=\frac{1}{2} m v^2$

$ \begin{aligned} & =\frac{1}{2} m \cdot(0.8 c)^2=0.32 m c^2 \\ \therefore \quad \frac{E}{E^{\prime}} & =\frac{h c / \lambda}{0.32 m c^2}=\frac{h}{\lambda(0.32 m c)} \\ & =\frac{h}{\frac{h}{m \times 0.8 c} \times 0.32 m c}(\text { from Eq. }(\mathrm{i})) \\ & =\frac{0.8}{0.32}=\frac{5}{2} \text { or }=5: 2 \end{aligned} $

Given:

Initial wavelength: $\lambda_1 = 1 \text{ nm}$

Final wavelength: $\lambda_2 = 0.5 \text{ nm}$

According to de-Broglie's principle, the wavelength ($\lambda$) is given by:

$ \lambda = \frac{h}{p} $

where $p$ (momentum) is related to the kinetic energy ($K$) by:

$ p = \sqrt{2mK} $

This leads to the expression for the kinetic energy in terms of wavelength:

$ K = \frac{h^2}{2m\lambda^2} $

where:

$p$ = momentum

$K$ = kinetic energy

$m$ = mass of the electron

$h$ = Planck's constant

Initial Kinetic Energy:

$ K_i = \frac{h^2}{2m(\lambda_1)^2} = \frac{h^2}{2m(1 \text{ nm})^2} $

Final Kinetic Energy:

$ K_f = \frac{h^2}{2m(\lambda_2)^2} = \frac{h^2}{2m(0.5 \text{ nm})^2} $

Change in Kinetic Energy:

The change in kinetic energy $\Delta K$ is calculated as:

$ \Delta K = K_f - K_i = \frac{h^2}{2m} \left(\frac{1}{(0.5)^2} - \frac{1}{1^2}\right) $

Simplifying, we get:

$ \Delta K = 3 \left(\frac{h^2}{2m}\right) = 3K_i $

Thus, the additional energy required to halve the wavelength is three times the initial energy.

The de Broglie wavelength of an electron, which is accelerated by a potential difference $ V $, is calculated using the following formula:

$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E_k}} $

where $ E_k = eV $ (with $ e $ being the charge of the electron and $ V $ the potential difference). This leads to:

$ \lambda = \frac{h}{\sqrt{2 m e V}} $

By substituting the known values of $ h $, $ m $, and $ e $ into the formula, we simplify it to:

$ \lambda = \frac{1.23}{\sqrt{V}} \text{ nm} $

To find the de Broglie wavelength for a potential difference of 900 V, we calculate:

$ \lambda = \frac{1.23}{\sqrt{900}} \text{ nm} = \frac{1.23}{30} \text{ nm} $

Thus, the de Broglie wavelength is approximately:

$ \lambda = 0.04 \text{ nm} $

Given, $v_{\max }=1.6 \times 10^6 \mathrm{~m} / \mathrm{s}$

$ e=1.6 \times 10^{-19} \mathrm{C}, m=9 \times 10^{-31} \mathrm{~kg} $

If $V_0$ be the stopping potential, then according to Einstein's photoelectric equation,

$ \begin{aligned} e V_0 & =\frac{1}{2} m v_{\max }^2 \\ \Rightarrow \quad V_0 & =\frac{m v_{\max }^2}{2 e} \\ & =\frac{9 \times 10^{-31} \times\left(1.6 \times 10^6\right)^2}{2 \times 1.6 \times 10^{-19}}=7.2 \mathrm{~V} \end{aligned} $

Photon flux = No. of photon reaching surface per unit area per unit time

$ \begin{aligned} & =\frac{\text { No. of photon emitted per sec }}{\text { Area of incidence }} \\ & =\frac{P}{E_1 \times A}=\frac{P}{\frac{h c}{\lambda} \times A} \\ & =\frac{942 \times 660 \times 10^{-9}}{\left(6.6 \times 10^{-34} \times 3 \times 10^8 \times 4 \times \pi \times 5^2\right)} \\ & \approx 1 \times 10^{19} \end{aligned} $

By increasing the potential difference between cathode and anode continuously in a photoelectric experiment, the photocurrent increases only upto certain value and after that remains constant (saturate).

Hence, statement I is false.

According to given situation is statement II,

$ \begin{aligned} \frac{\left(K_{\max }\right)_A}{\left(K_{\max }\right)_B} & =\frac{E_A-\phi}{E_B-\phi} \\ & =\frac{2.5-2}{3.5-2}=\frac{0.5}{1.5}=\frac{1}{3} \end{aligned} $

Hence, statement II is also false.

The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal surface.

Hence, statement III is also false.

de-Broglie wavelength of a moving particle is given as

$ \begin{array}{ll} \lambda =\frac{h}{m v} \\ \Rightarrow \quad \lambda \propto \frac{1}{m v} \end{array} $

Since, product of $m v$ for a ball of mass 0.03 kg moving with velocity $1 \mathrm{~m} / \mathrm{s} =1 \times 0.03=0.03 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ which is minimum among the all given options. Therefore, de-Broglie wavelength of ball of mass 0.03 kg moving with velocity $1 \mathrm{~m} / \mathrm{s}$ is largest.

According to Einstein's photoelectric equation,

$ e V_0=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) $

when, $\lambda=\lambda_1=200 \mathrm{~nm}$, then

$ e\left(V_0\right)_1=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_0}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

when, $\lambda=\lambda_2=400 \mathrm{~nm}$

$ \therefore e\left(V_0\right)_2=h c\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_0}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∴ Subtracting Eq. (ii) form Eq. (i),

$ \begin{aligned} & e\left(V_0\right)_1-e\left(V_0\right)_2=h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_0}-\frac{1}{\lambda_2}+\frac{1}{\lambda_0}\right] \\ & \begin{aligned} \Rightarrow e\left[\left(V_0\right)_1-\left(V_0\right)_2\right] & =h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \\ e\left[\left(V_0\right)_1-\left(V_0\right)_2\right] & =1240 \mathrm{eV} \\ - & \mathrm{nm}\left[\frac{1}{200 \mathrm{~nm}}-\frac{1}{400 \mathrm{~nm}}\right] \\ \Rightarrow\left(V_0\right)_1-\left(V_0\right)_2 & =1240\left[\frac{2-1}{400}\right] \\ = & 1240 \times \frac{1}{400}=3.1 \mathrm{~V} \end{aligned} \end{aligned} $

Given, kinetic energy of electron,

$ \begin{aligned} K & =320 \mathrm{eV} \\ & =320 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & =512 \times 10^{-17} \mathrm{~J} \\ h & =6 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned} $

Mass of electron,

$ \begin{aligned} m_e & =9 \times 10^{-31} \mathrm{~kg} \\ e & =1.6 \times 10^{-19} \mathrm{C} \end{aligned} $

∴ de-Broglie wavelength of electron,

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 \mathrm{mK}}} \\ & =\frac{6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 512 \times 10^{-17}}} \\ & =\frac{6 \times 10^{-34}}{9.6 \times 10^{-24}} \\ & =0.625 \times 10^{-10} \mathrm{~m} \\ & =62.5 \times 10^{-12} \mathrm{~m} \\ & =62.5 \mathrm{pm}\end{aligned} $

$ \quad\left[\because 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right] $

Given, wavelength of incident radiation, $\lambda=248 \mathrm{~nm}=248 \times 10^{-7} \mathrm{~m}$

Stopping potential, $V_0=2.8 \mathrm{eV}$ According to Einstein's equation,

$ \begin{aligned} e V_0 & =h v-\phi_0 \\ \Rightarrow \quad e V_0 & =\frac{h c}{\lambda}-\phi_0 \\ \Rightarrow \quad \phi_0 & =\frac{h c}{\lambda}-e V_0 \\ & =\frac{1240}{248}-2.8=5-2.8 \\ & =2.2 \mathrm{eV} \end{aligned} $

Given, potential difference,

$ \begin{aligned} & V=121 \mathrm{~V} \\ & h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}, m=9 \times 10^{-31} \mathrm{~kg} \end{aligned} $

de-Broglie wavelength is given as

$ \begin{aligned} \lambda & =\frac{h}{\sqrt{2 \mathrm{meV}}} \\ & =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 121}} \\ & =0.112 \times 10^{-9} \mathrm{~m} \\ & =0.112 \mathrm{~nm} \end{aligned} $

Given, slope of graph between stopping potential versus frequency is equal to $\frac{h}{e}$.

where, $h$ is planck's constant.

$ \begin{array}{ll} \therefore \frac{h}{e}=\text { slope }=4 \times 10^{-15} \\ \Rightarrow \quad h=e \times 4 \times 10^{-15} \\ =1.6 \times 10^{-19} \times 4 \times 10^{-15}=6.4 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{array} $

In photoelectric effect, kinetic energy of emitted photoelectrons is directly proportional to the frequency of emitted photoelectrons. Hence, on increasing frequency of incident radiation, kinetic energy of emitted photoelectron increases. Hence, statement I is incorrect and statement IV is correct.

In photoelectric effect, photoelectric emission takes place only when frequency of incident radiation is greater than or equal to threshold frequency, no matter how much intensity of incident light is. Hence, statement II is not correct.

In photoelectric emission, maximum kinetic energy of emitted photoelectrons depends only frequency of incident radiation and does not depend on intensity of incident radiation. Hence, statement III is correct.

Photon energy is dependent on the frequency or wavelength of radiation. It does not depend on the intensity of radiation.

Since, $E=h v=\frac{h c}{\lambda}$

i.e. $E \propto v$ or $E \propto \frac{1}{\lambda}$

where, $\mathrm{v}=$ frequency and

$\lambda=$ wavelength.

Given, transition levels,

$ n_1=2, n_2=3 $

Energy emitted during transition

$ n_2=3 \text { to } n_1=2 \text { in H-atom } $

(i.e. for $\mathrm{H}_\alpha$ - light)

$ \begin{aligned} E & =13.6\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ & =13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =13.6\left[\frac{5}{36}\right] \mathrm{eV}=1.89 \mathrm{eV} \end{aligned} $

For $\mathrm{H}_\alpha$-light to be able to emit photoelectrons from a metal, the work function must be less than or equal to 1.89 eV .

Hence, maximum value of work function $=1.89 \mathrm{eV}$

The energy equation in photoelectric effect is given as

$ \begin{array}{rlrl} e V_0 & =h v-\phi_0 \\ \Rightarrow \quad & V_0 =\frac{h v}{e}-\frac{\phi_0}{e} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

When the frequency is doubled,

$ \begin{aligned} V_0^{\prime} & =\frac{2 h v}{e}-\frac{\phi_0}{e} \\ \Rightarrow \quad V_0^{\prime} & =2\left(\frac{h v}{e}-\frac{\phi_0}{e}\right)+\frac{\phi_0}{e} \\ V_0^{\prime} & =2 V_0+\frac{\phi_0}{e} \quad \text { [using Eq. (i)] } \end{aligned} $

$ \Rightarrow \quad V_0^{\prime}>2 V_0 $

Thus, the stopping potential will become more than double.

Given, work-function, $\phi=2.4 \mathrm{eV}$

Energy required to ionise H -atom in ground state $=13.6 \mathrm{eV}$

According to photoelectric equation,

$ \begin{aligned} E & =\mathrm{KE}+\phi \Rightarrow \frac{h c}{\lambda}=\mathrm{KE}+\phi \\ \Rightarrow \quad \frac{1240}{\lambda} & =13.6+2.4 \\ & (\because h c=1240 \mathrm{eV}-\mathrm{nm}) \end{aligned} $

$ \begin{aligned} \Rightarrow & \frac{1240}{\lambda} =16 \\ \therefore & \lambda =\frac{1240}{16}=77.5 \mathrm{~nm} \end{aligned} $

Work-function of metal surface, $\phi_0=25 \mathrm{eV}$, Wavelength, $\lambda=310 \mathrm{~nm}$

$ \Rightarrow h c=1240 \mathrm{eV}-\mathrm{nm} $

According to Einstein's photoelectric equation,

$ \begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi_0 \\ \Rightarrow \quad e V_0 & =\frac{1240 \mathrm{eV}-\mathrm{nm}}{310 \mathrm{~nm}}-2.5 \mathrm{eV} \\ \Rightarrow \quad V_0 & =\frac{1240}{310}-25=4-25=1.5 \mathrm{~V} \end{aligned} $

In case of photoemission,

$ K_{\max }=\frac{1}{2} m v_{\max }^2=h f-\phi_0 $

For photons of energies 4 eV and 2.5 eV , we have (given, work function $=2 \mathrm{eV}$ )

$ \frac{1}{2} m v_1^2=4-2=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\quad \frac{1}{2} m v_2^2=2.5-2=0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (i) by Eq. (ii), we get

$ \frac{v_1^2}{v_2^2}=\frac{2}{0.5}=4 \Rightarrow \frac{v_1}{v_2}=2 $

$ \text { Let } \phi_0=\text { work-function of metal. } $

Then, according to Einstein's photoelectric equation,

$ K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When wavelength is reduced by $50 \%$ i.e, $\lambda^{\prime}=\frac{\lambda}{2}$, then maximum velocity of emitted electrons is $3 v$.

$ \begin{array}{ll} \Rightarrow & \frac{1}{2} m(3 v)^2=\frac{h c}{\lambda / 2}-\phi_0 \\ \Rightarrow & 9\left(\frac{1}{2} m v^2\right)=\frac{2 h c}{\lambda}-\phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & 9\left(\frac{h c}{\lambda}-\phi_0\right)=\frac{2 h c}{\lambda}-\phi_0 \\ & 7 \frac{h c}{\lambda}=8 \phi_0 \Rightarrow \phi_0=\frac{7}{8} \frac{h c}{\lambda} \end{aligned} $

Here, photon energy is given as

$ \frac{h c}{\lambda}=2.4 \mathrm{eV} $

So, work-function is

$ \phi_0=\frac{7}{8} \times 2.4=2.1 \mathrm{eV} $