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The Photoelectric Effect MCQ - Practice Questions with Answers

The Photoelectric Effect MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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Quick Facts

Photoelectric effect is considered one the most difficult concept.
Graphs in Photoelectric effect is considered one of the most asked concept.
88 Questions around this concept.

Solve by difficulty

Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from the choices given below the list :

A

(A) - (i) (B) - (iv) (C) - (iii)

Correct Answer

Wrong Answer

B

(A) - (ii) (B) - (iv) (C) - (iii)

Correct Answer

Wrong Answer

C

(A) - (ii) (B) - (i) (C) - (iii)

Correct Answer

Wrong Answer

D

(A) -(iv) (B) - (iii) (C) - (ii)

Correct Answer

Wrong Answer

When photons of wavelength λ₁ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ₂ are used, the corresponding stopping potential is thrice that of the above value. If light of wavelength λ₃ is used then find the stopping potential for this case :

A

$\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} - \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]$

Correct Answer

Wrong Answer

B

$\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]$

Correct Answer

Wrong Answer

C

$\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{3}{2\lambda _{1}}\right ]$

Correct Answer

Wrong Answer

D

$\frac{hc}{e}\left [ \frac{1}{\lambda _{3}} + \frac{1}{2\lambda _{2}} - \frac{1}{\lambda _{1}}\right ]$

Correct Answer

Wrong Answer

Question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

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 false.

Correct Answer

Wrong Answer

B

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Correct Answer

Wrong Answer

C

Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

Correct Answer

Wrong Answer

D

Statement-1 is false, Statement-2 is true.

Correct Answer

Wrong Answer

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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 $\upsilon > \upsilon _{0}$ (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 $K_{max}\;$ and $V_{0}$ 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 false.

Correct Answer

Wrong Answer

B

Statement-1 is true , Statement-2 is true, Statement-2 is the correct explanation of Statement-1.

Correct Answer

Wrong Answer

C

Statement-1 is true , Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

Correct Answer

Wrong Answer

D

Statement-1 is false, Statement-2 is true.

Correct Answer

Wrong Answer

Two identical photo cathodes receive light of frequencies $f_{1}\, and\, f_{2}$ . If the velocities of the photoelectrons (of mass m ) coming out are respectively $v_{1}\, and\, v_{2}$ , then

A

$v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})$

Correct Answer

Wrong Answer

B

$v_{1}+v_{2}=\left [ \frac{2h}{m}(f_{1}+f_{2}) \right ]^{\frac{1}{2}}$

Correct Answer

Wrong Answer

C

$v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}(f_{1}+f_{2})$

Correct Answer

Wrong Answer

D

$v_{1}-v_{2}=\left [ \frac{2h}{m}(f_{1}-f_{2}) \right ]^{\frac{1}{2}}$

Correct Answer

Wrong Answer

According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal and the frequency of the incident radiation gives a straight line whose slope is:

A

depends on the nature of the metal used

Correct Answer

Wrong Answer

B

depends on the intensity of the radiation

Correct Answer

Wrong Answer

C

depends both on the intensity of the radiation and the metal used

Correct Answer

Wrong Answer

D

is the same for all metals and independent of the intensity of the radiation.

Correct Answer

Wrong Answer

Photoelectric emission occurs only when the incident light has more than a certain minimum

A

power

Correct Answer

Wrong Answer

B

wavelength

Correct Answer

Wrong Answer

C

intensity

Correct Answer

Wrong Answer

D

frequency

Correct Answer

Wrong Answer

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For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

A

$\sqrt{\frac{hv}{m}}$

Correct Answer

Wrong Answer

B

$\sqrt{\frac{2hv}{m}}$

Correct Answer

Wrong Answer

C

$2\sqrt{\frac{hv}{m}}$

Correct Answer

Wrong Answer

D

$\sqrt{\frac{hv}{(2m)}}$

Correct Answer

Wrong Answer

Light of two different frequencies whose photons have energies 1 eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be

A

1:4

Correct Answer

Wrong Answer

B

1:2

Correct Answer

Wrong Answer

C

1:1

Correct Answer

Wrong Answer

D

1:5

Correct Answer

Wrong Answer

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The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

A

2.4 V

Correct Answer

Wrong Answer

B

-1.2 V

Correct Answer

Wrong Answer

C

-2.4 V

Correct Answer

Wrong Answer

D

1.2 V

Correct Answer

Wrong Answer

Concepts Covered - 2

Photoelectric effect

Photoelectric effect-

The phenomea of Photoelectric effect was first introduced by Wilhelm Ludwig Franz Hallwachs in 1887 and its experimental verification was confirmed by Heinrich Rudolf Hertz. They observed that when a metallic surface is irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomena of ejection of electron is called the Photoelectric effect. Now question arrise that what is Photoelectric effect??

The photoelectric effect is the process that involves the release or rejection of electrons from the surface of materials (this material is generally a metal) when light falls on them. This concept thatmakes us comfortable to understand quantum nature of electron and light.

The electrons ejected during photoelectric effect were called as photoelectrons.There is one condition for photoelectric effect which is very much important that for photoemission to take place, energy of incident light photons should be greater than or equal to the work function of the metal. Now what is work function????

Work function $(\phi)$ is defined as the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid, usually a metal.

Now on the basis of work function $(\phi)$ we can define two related quantity which are Threshold frequency and Threshold wavelength. Now as we know that the energy is of photon is given by -

$E = h \nu = \frac{hc}{\lambda}$

Now the frequency corresponding to the energy equals to work function is called Threshold frequency and similarly the wavelength corresponding to the work function is Threshold wavelength.

$\phi=h \nu_{t h} \quad \nu_{t h}=\frac{\phi}{h}$

$Similarly \ \ \phi=\frac{hc}{\lambda _{th}} \quad \lambda_{t h}=\frac{hc}{\phi}$

Now, let us understand with experiment which was performed during Heinrich Rudolf Hertz. For this let us consider the given set-up-

In this experiment set-up, an evacuated glass tube is there. Two zinc plates C and A are enclosed. Plates A acts as anode and C acts as a photosensitive plate. Two plates are connected to a battery and ammeter as shown. If the radiation is incident on the plate C through a quartz window, electrons are ejected out of plate and current flows in the circuit this is known as photocurrent. Plate A can be maintained at desired potential (+ve or – ve) with respect to plate C.

Applications of Photoelectric Effect -

This phenomena is used to generate electricity in Solar Panels.
We come across many sensor in our day to day life. Few sensors are also working in Photoelectric effect.
It is also used in digital cameras because they have photoelectric sensor.

Note -

In case of Threshold frequency - If incident frequency $v< \nu _{0}$ . No photoelectron emission. The minimum frequency of incident radiation to eject electron is threshold frequency ( $\nu_{0}$ )
In case of Threshold Wavelength - If $\lambda > \lambda _{0}$ No photoelectron emission. The maximum wave length of incident raditaion required to eject the electron is Threshold Wavelength $(\lambda_{0})$
Work function
$h= Planck's \: constant$

$\nu _{0}= threshold \: frequency$

Energy used to overcome the surface barrier and come out of metal surface.

$\phi = h\nu _{0}$
Kinetic Energy of Photo eletrons

$m\rightarrow mass\: of\: photoelectron$

Remaining part of the energy is used in gaining a velocity $v$ to the emitted photoelectron

$k_{max}= \frac{1}{2}mv^{2}_{max}$
Conservation of energy

$h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}$

$h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}$

$h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}$

$where, h - Planck's \ constant,\ \nu - Frequency,\ \nu_{0} - threshold\ frequency, \ \phi_{0}- work \ function$

Graphs in Photoelectric effect

Graphs related to Photoelectric effect -

Before giving the variation in the graph we should define some important terminologies which are used while plotting the graph.

1. Stopping potential-

The negative potential of the collector plate at which the photoelectric current becomes zero is called stopping potential or cut-off potential. Stopping potential is that value of retarding potential difference between two plates which is just sufficient to stop the most energetic photoelectrons emitted. It is denoted by " $V_o$ ".

We need to equate maximum kinetic energy Kmax of the photo-electron (having charge e) to the stopping potential V_o

We know that,
Electric potential energy=Potential Difference×Charge
So,

$U=V_{0} \times Q$

$\begin{array}{l}{U=K_{\max }} \\ {\therefore\left|V_{0} \times e\right|=K_{\max }} \\ {\Rightarrow K_{\max }=\left|e V_{0}\right|}\end{array}$

By using previous concept, we can write that -

$h \nu = h \nu_o + K.E._{(max)}$

Now since K.E._max = eV_o, So we can write that -

$\begin{array}{l}{\mathrm{eV}_{\mathrm{s}}=\mathrm{h}\left(\mathrm{\nu}-\mathrm{\nu}_{\mathrm{o}}\right)} \\ {\text { or }} \\ {\mathrm{V}_{\mathrm{s}}=\frac{\mathrm{h}}{\mathrm{e}}\left(\mathrm{\nu}-\mathrm{\nu}_{\mathrm{o}}\right)}\end{array}$

The above graph shows the variation between the stopping potential and frequency.

2. Saturation current -

The photoelectric current attains a saturation value and does not increase further for any increase in the positive potential. It means that this photoelectric current is the saturation current even we are increasing the value of the positive potential.

Now let us discuss the variation one by one in detail -

1. Variation of photocurrent with intensity -

2. Variation of photoelectric current with potential and intensity

3. Effects of frequency of incident light on the stopping potential -

4. Variation of Kinetic energy with frequency

Study it with Videos

Photoelectric effect

Graphs in Photoelectric effect

Study other Related Concepts

The Photoelectric Effect Current Topic

Electron Emission

Photon Theory Of Light

The Photoelectric Effect

Einstein's Photoelectric Equation

Radiation Pressure

Wave Nature Of Matter And De Broglie's Equation

Davisson-germer Experiment

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