iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Morning Slot
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 n = ${{1240} \over {\lambda (in\,mm)}}$eV)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths $\lambda $N, $\lambda $A respectively. The ratio ${{{}^\lambda N} \over {{}^\lambda A}}$is closest to :
A.
10$-$6
B.
10
C.
10$-$10
D.
10$-$1
Correct Answer: A
Explanation:
We know that $E = {{hc} \over \lambda }$
So, for atom ${E_A} = {{hc} \over {{\lambda _A}}}$
And for neutron ${E_N} = {{hc} \over {{\lambda _N}}}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
The de-Broglie wavelength ($\lambda $B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($\lambda $G) by :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Evening Slot
If the de Broglie wavelengths associated with a proton and an $\alpha $-particle are equal, then the ratio of velocities of the proton and the $\alpha $-particle will be :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 15th April Morning Slot
Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are ${\lambda _1}$ and ${\lambda _2},$ their de Broglie wavelength in the frame of reference attached to their center of masses :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 9th April Morning Slot
A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
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 :
A.
less than $\sqrt 3 $ v
B.
v
C.
more than $\sqrt 3 \,v$
D.
equal to $\sqrt 3 \,v$
Correct Answer: C
Explanation:
The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by
$KE = {1 \over 2}m{v^2} = hn - \phi $
where h is Planck's constant and $\phi$ is the work function. Therefore, the kinetic energy of the incident light is
${E_1} = hn - \phi $ ..... (1)
When the frequency of the incident light is increased to 3n, then the kinetic energy is given by
${1 \over 2}mv_1^2 = 3hn - \phi $
$ \Rightarrow {E_2} = 3hn - \phi $ ..... (2)
Substituting $hn = {E_1} + \phi $ [from Eq. (1)] in Eq. (2), we get
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
An electron beam is accelerated by a potential difference V to hit a metallic target to produce X–rays. It
produces continuous as well as characteristic X-rays. If $\lambda $min is the smallest possible wavelength of X-ray in the spectrum, the variation of log$\lambda $min with log V is correctly represented in:
A.
B.
C.
D.
Correct Answer: B
Explanation:
In X-ray tube, ${\lambda _{\min }} = {{hc} \over {eV}}$
Clearly, log ($\lambda $min) versus log V graph
slope is negative hence option (b) is correct.
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The
collision is head on, and elastic. The ratio of the de-Broglie wavelengths ${\lambda _A}$ to ${\lambda _B}$ after the collision is:
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 10th April Morning Slot
A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda $ and ${\lambda \over 2}.$ 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 9th April Morning Slot
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Offline)
Radiation of wavelength $\lambda ,$ is incident on a photocell. The fastest emitted electron has speed $v.$ If the wavelength is changed to ${{3\lambda } \over 4},$ the speed of the fastest emitted electron will be:
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2015 (Offline)
Match List - ${\rm I}$ (Fundamental Experiment) with List - ${\rm II}$ (its conclusion) and select the correct option from the choices given below the list:
A.
$A - ii;\,\,B - i,\,\,C - iii$
B.
$A - iv;\,\,B - iii,\,\,C - ii$
C.
$A - i;\,\,B - iv,\,\,C - iii$
D.
$A - ii;\,\,B - iv,\,\,C - iii$
Correct Answer: A
Explanation:
Frank-Hertz experiment - Discrete energy levels of atom
Photoelectric effects - Particle nature of light
Davison - Germer experiment - wave nature of electron.
2013
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2013 (Offline)
The anode voltage of a photocell is kept fixed. The wavelength $\lambda $ of the light falling on the cathode is gradually changed. The plate current $I$ of the photocell varies as follows :
A.
B.
C.
D.
Correct Answer: D
Explanation:
As $\lambda $ is increased, there will be a value of $\lambda $ above which photo electrons will be cases to come out so photo current will become zero. Hence $(d)$ is correct answer.
This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement 1 : Davisson - Germer experiment established the wave nature of electrons.
Statement 2 : If electrons have wave nature, they can interfere and show diffraction.
A.
Statement 1 is true, Statement 2 is false
B.
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1.
C.
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1.
D.
Statement 1 is false, Statement 2 is true.
Correct Answer: B
Explanation:
Davisson-Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction.
This question has Statement - $1$ and Statement - $2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement - $1$ : A metallic surface is irradiated by a monochromatic light of frequency $v > {v_0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are ${K_{\max }}$ and ${V_0}$ respectively. If the frequency incident on the surface is doubled, both the ${K_{\max }}$ anmd ${V_0}$ are also doubled.
Statement - $2$ : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
A.
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is the correct explanation of Statement - $1$.
B.
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is not the correct explanation of Statement - $1$.
C.
Statement - $1$ is false, Statement - $2$ is true.
D.
Statement - $1$ is true, Statement - $2$ is false.
Correct Answer: C
Explanation:
By Einstein photoelectric equation,
${K_{\max }} = e{V_0} = hv - h{v_0}$
When $v$ is doubled, ${K_{\max }}$ and ${V_0}$ become more than double.
Statement - $1$ : When ultraviolet light is incident on a photocell, its stopping potential is ${V_0}$ and the maximum kinetic energy of the photoelectrons is ${K_{\max }}$. When the ultraviolet light is replaced by $X$-rays, both ${V_0}$ and ${K_{\max }}$ increase.
Statement - $2$ : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
A.
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is the correct explanation of Statement - $1$
B.
Statement - $1$ is true, Statement - $2$ is true; Statement - $2$ is not the correct explanation of Statement - $1$
C.
Statement - $1$ is is false, Statement - $2$ is true
D.
Statement - $1$ is is true, Statement - $2$ is false
Correct Answer: B
Explanation:
Statement 1 is true. The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the light. X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the photoelectrons. This results in a higher stopping potential ($V_0$) and maximum kinetic energy ($K_{\max}$) for the photoelectrons.
Statement 2 is also true. However, while the speeds (and hence kinetic energies) of photoelectrons do vary, this variation is not because of a range of frequencies in the incident light. Rather, it's due to the interaction of the incident photons with electrons at different energy levels in the metal. A single frequency of light can produce photoelectrons with a range of speeds because the electrons they encounter can have a variety of binding energies.
The surface of a metal is illuminated 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 : $\left( {hc = 1240eV.nm} \right)$
A.
$1.41$ $eV$
B.
$1.51$ $eV$
C.
$1.68$ $eV$
D.
$3.09$ $eV$
Correct Answer: A
Explanation:
The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by:
$E = K_{\max} + W$,
where
$E$ is the energy of the incident light,
$K_{\max}$ is the maximum kinetic energy of the photoelectrons, and
$W$ is the work function of the metal.
The energy of the incident light can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. However, given that $hc = 1240$ eV⋅nm, we can simplify this to $E = \frac{1240}{\lambda}$.
Substituting the given values into this equation, we have:
$E = \frac{1240}{400} = 3.1$ eV.
We can then substitute these values into the photoelectric effect equation:
$3.1 \text{ eV} = 1.68 \text{ eV} + W$,
which simplifies to:
$W = 3.1 \text{ eV} - 1.68 \text{ eV} = 1.42$ eV.
Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.
Thus, Option A: $1.41$ eV is the closest to the correct answer.
In an experiment, electrons are made to pass through a narrow slit of width $'d'$ comparable to their de Broglie wavelength. They are detected on a screen at a distance $'D'$ from the slit (see figure).
Which of the following graphs can be expected to represent the number of electrons $'N'$ detected as a function of the detector position $'y'\left( {y = 0} \right.$ corresponds to the middle of the slit$\left. \, \right)$
A.
B.
C.
D.
Correct Answer: D
Explanation:
The electron beam will be diffracted and the maxima is obtained at $y=0.$ Also the distance between the first minima on both side will be greater than $d.$
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).
If a strong diffraction peak is observed when electrons are incident at an angle $'i'$ from the normal to the crystal planes with distance $'d'$ between them (see figure), de Broglie wavelength ${\lambda _{dB}}$ of electrons can be calculated by the relationship ($n$ is an integer)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).
Electrons accelerated by potential $V$ are diffracted from a crystal. If $d = 1\mathop A\limits^ \circ $ and $i = {30^ \circ },\,\,\,V$ should be about
$\left( {h = 6.6 \times {{10}^{ - 34}}Js,{m_e} = 9.1 \times {{10}^{ - 31}}kg,\,e = 1.6 \times {{10}^{ - 19}}C} \right)$
A.
$2000$ $V$
B.
$50$ $V$
C.
$500$ $V$
D.
$1000$ $V$
Correct Answer: B
Explanation:
Using Bragg's equation $2d$ $\sin \theta = n\lambda $
Here $n=1,$ $\theta = 90 - i = 90 - 30 = {60^ \circ }$
$\therefore$ $2d$ $\sin \,\theta \, = \lambda \,\,\,\,\,\,\,\,\,\,....\left( i \right)$
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 $5V.$ The incident radiation lies in
A.
ultra-violet region
B.
infra-red region
C.
visible region
D.
$x$-ray region
Correct Answer: A
Explanation:
The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation
$E = eV,$
where e is the elementary charge.
The elementary charge, e, is approximately equal to $1.602 \times 10^{-19}$ C. So the energy of the incident radiation is
This is less than the threshold energy of $6.2 \, \text{eV},$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.
The regions of the electromagnetic spectrum are generally classified by energy as follows:
Infrared: Less than about $1.24 \, \text{eV}$
Visible: Between about $1.24 \, \text{eV}$ and $3.1 \, \text{eV}$
Ultraviolet: Between about $3.1 \, \text{eV}$ and $124 \, \text{eV}$
X-rays: Greater than about $124 \, \text{eV}$
Therefore, the incident radiation falls in the ultraviolet region.
The anode voltage of a photocell is kept fixed. The wavelength $\lambda $ of the light falling on the cathode is gradually changed. The plate current ${\rm I}$ of the photocell varies as follows
A.
B.
C.
D.
Correct Answer: B
Explanation:
As $\lambda $ decreases, $v$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.
A photocell is illuminated by a small bright source placed $1$ $m$ away. When the same source of light is placed ${1 \over 2}$ $m$ away, the number of electrons emitted by photo-cathode would
When intensity becomes 4 times, no. of photoelectrons emitted would increase by $4$ times, since number of electrons emitted per second is directly proportional to intensity.
The work function of a substance is $4.0$ $eV.$ The longest wavelength of light that can cause photo-electron emission from this substance is approximately.
According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $Vs$ the frequency, of the incident radiation gives as straight the whose slope
A.
depends both on the intensity of the radiation and the metal used
B.
depends on the intensity of the radiation
C.
depends on the nature of the metal used
D.
is the same for the all metals and independent of the intensity of the radiation
Correct Answer: D
Explanation:
Einstein's photoelectric equation is given by
$K_{\max} = h\nu - \phi,$
where $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident radiation, and $\phi$ is the work function of the metal (the minimum energy needed to eject an electron).
If you plot $K_{\max}$ against $\nu$, you will get a straight line with slope $h$ (Planck's constant) and y-intercept $-\phi$ (the negative of the work function). The slope of the line (which is $h$) does not depend on the intensity of the radiation or the type of metal used. Instead, it is a universal constant.
Therefore, the correct answer is:
Option D: The slope is the same for all metals and independent of the intensity of the radiation.
Two identical photo-cathodes receive light of frequencies ${f_1}$ and ${f_2}$. If the velocities of the photo electrons (of mass $m$ ) coming out are respectively ${v_1}$ and ${v_2},$ then
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2026 (Online) 28th January Morning Shift
The ratio of de Broglie wavelength of a deutron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2 E$, is $n: 1$. The value of $n$ is $\_\_\_\_$ .
(Assume mass of proton $=$ mass of neutron) $:$
Correct Answer: 2
Explanation:
The de Broglie wavelength is given by :
$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $
Since kinetic energy $\mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$, we can write momentum as $\mathrm{p}=\sqrt{2 \mathrm{mK}}$.
So, the ratio of de-Broglie wavelengths is $2: 1 \Rightarrow \frac{\mathrm{n}}{1}=\frac{2}{1} \Rightarrow \mathrm{n}=2$
Therefore, the value of $n$ is 2 . Hence, the correct answer is 2.
2026
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2026 (Online) 21st January Evening Shift
A particle having electric charge $3 \times 10^{-19} \mathrm{C}$ and mass $6 \times 10^{-27} \mathrm{~kg}$ is accelerated by applying an electric potential of 1.21 V .
Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12} \mathrm{~m}$. The value of $\alpha$ is $\_\_\_\_$ .
When a particle of charge $q$ is accelerated through a potential difference $V$, the work done on the particle is converted into its kinetic energy (K):
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 8th April Evening Shift
An electron is released from rest near an infinite non-conducting sheet of uniform charge density '$-\sigma$'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical value of n is _____.
Correct Answer: 2
Explanation:
Let the momentum of $\mathrm{e}^{-}$at any time t is p and its de-broglie wavelength is $\lambda$.
Then, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$
$\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}$
Where, -ve sign represents decrease in $\lambda$ with time
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 24th January Evening Shift
The ratio of the power of a light source $S_1$ to that the light source $S_2$ is $2 . S_1$ is emitting $2 \times 10^{15}$ photons per second at 600 nm . If the wavelength of the source $S_2$ is 300 nm , then the number of photons per second emitted by $S_2$ is __________ $\times 10^{14}$.
Correct Answer: 5
Explanation:
$\begin{aligned}
&\text { Since power emitting by a source is given as }\\
&\begin{aligned}
& =\frac{\text { Total energy emitted }}{\text { time }} \\
& =\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\
& P_1=\left(E_1\right) n
\end{aligned}
\end{aligned}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2024 (Online) 6th April Evening Shift
In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $10.2 \mathrm{~V}$. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given hc $=1245 \mathrm{~eV} \mathrm{~nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$).
Correct Answer: 122
Explanation:
The Franck-Hertz experiment provides evidence for quantized energy levels within atoms. When atoms are excited by electrons with a specific kinetic energy, they can jump to higher energy levels. Upon returning to lower levels, they emit photons whose energies correspond to the difference between these levels. The first dip in the current-voltage graph for hydrogen, observed at $10.2 \mathrm{~V}$, corresponds to the energy required to excite a hydrogen atom to its first excitation level. The wavelength of the light emitted when the atom returns to its ground state can be calculated using the energy of the photon emitted.
To find the wavelength ($\lambda$) of light emitted, we use the relationship between energy ($E$), Planck's constant ($h$), the speed of light ($c$), and wavelength ($\lambda$), given in the equation form as $E = \frac{hc}{\lambda}$.
However, we are given $hc$ in electron volts per nanometer ($1245 \mathrm{~eV} \cdot \mathrm{nm}$), and the energy is also given in terms of voltage ($10.2 \mathrm{~V}$). First, we convert the energy into electron volts (eV) using the formula: $E = eV$, where $e$ is the charge of an electron ($1.6 \times 10^{-19} \mathrm{C}$).
The energy in electron volts can be directly calculated as:
Therefore, the wavelength of light emitted by the hydrogen atom when excited to the first excitation level is approximately $122.06 \mathrm{~nm}$.
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 13th April Evening Shift
An atom absorbs a photon of wavelength $500 \mathrm{~nm}$ and emits another photon of wavelength $600 \mathrm{~nm}$. The net energy absorbed by the atom in this process is $n \times 10^{-4} ~\mathrm{eV}$. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take $\mathrm{h}=6.6 \times 10^{-34} ~\mathrm{Js}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ )
Correct Answer: 4125
Explanation:
The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula:
$E=\frac{hc}{\lambda}$
where $h$ is Planck's constant and $c$ is the speed of light.
In this problem, we are given that an atom absorbs a photon of wavelength $\lambda_1=500~\mathrm{nm}$ and emits another photon of wavelength $\lambda_2=600~\mathrm{nm}$. We can use the formula above to calculate the energy absorbed by the atom:
We need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron:
Finally, we can express the net energy absorbed in terms of the given value of $n$, as follows:
$n\times10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}$
Solving for $n$, we get:
$n=\frac{0.4125}{10^{-4}}=4125$
Therefore, the value of $n$ is $\boxed{4125}$.
2023
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2023 (Online) 11th April Morning Shift
A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $x \times 10^{15} \mathrm{~Hz}$. The value of $x$ is ____________.
(Given h $=4.25 \times 10^{-15} ~\mathrm{eVs}$ )
Correct Answer: 3
Explanation:
When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.
The energy levels of the hydrogen atom are given by the formula:
$E_n = -\frac{13.6 \mathrm{~eV}}{n^2}$
where $E_n$ is the energy of the nth level and $n$ is the principal quantum number.
Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.
The six transitions correspond to the following energy level changes:
From n = 2 to n = 1
From n = 3 to n = 1
From n = 3 to n = 2
From n = 4 to n = 1
From n = 4 to n = 2
From n = 4 to n = 3
The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 26th June Evening Shift
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\mathop A\limits^o $ is 0.42 V. If the threshold frequency is x $\times$ 1013 /s, where x is _________ (nearest integer).
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be __________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Morning Shift
A particle of mass 9.1 $\times$ 10$-$31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10$-$27 kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Evening Shift
A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ___________ $\times$ 10$-$7 T.
Given hc = 20 $\times$ 10$-$26 J-m, mass of electron = 9 $\times$ 10$-$31 kg
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Evening Shift
A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2$\lambda$, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ____________ $\lambda$.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th February Evening Shift
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 ___________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 5th September Evening Slot
The surface of a metal is illuminated alternately
with photons of energies E1 = 4 eV and E2 = 2.5 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 _____.
$ \therefore $ Work function($\phi $) of the metal = 2 eV
2020
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 5th September Morning Slot
A beam of electrons of energy E scatters from
a target having atomic spacing of 1 $\mathop A\limits^o $. The first
maximum intensity occurs at $\theta $ = 60o. Then E (in
eV) is ______.
(Planck constant h = 6.64 × 10–34 Js, 1 eV =
1.6 × 10–19 J, electron mass m = 9.1 × 10–31 kg)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 2nd September Morning Slot
When radiation of wavelength $\lambda $ is used to
illuminate a metallic surface, the stopping
potential is V. When the same surface is
illuminated with radiation of wavelength 3$\lambda $,
the stopping potential is
${V \over 4}$. If the threshold
wavelength for the metallic surface is n$\lambda $ then
value of n will be __________.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 7th January Morning Slot
A beam of electromagnetic radiation of intensity 6.4 × 10–5 W/cm2 is comprised of wavelength,
$\lambda $ = 310 nm. It falls normally on a metal (work function $\phi $ = 2eV) of surface area of 1 cm2. If one
in 103 photons ejects an elctron, total number of electrons ejected in 1 s is 10x. (hc = 1240 eVnm,
1eV = 1.6 × 10–19 J), then x is _____.
Correct Answer: 11
Explanation:
Energy of photon = ${{1240} \over {310}}$ = 4 eV
Energy is greater than work function so photoelectric effect will take place.
