Dual Nature Of Radiation And Matter | Physics JEE previous year questions with Solutions

Q 21 :

The work function of metal is 3 eV. The colour of the visible light that is required to cause emission of photoelectrons is [2025]

Green
Blue
Red
Yellow

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(2)

For emission : $\frac{h c}{λ} > ϕ$

$λ < \frac{h c}{ϕ} \Rightarrow λ < \frac{1242}{3} nm$

Q 22 :

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A) : In photoelectric effect, on increasing the intensity of incident light the stopping potential increases.

Reason (R) : Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.

In the light of the above statements, choose the correct answer from the options given below: {2025]

Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
(A) is false but (R) is true.
(A) is true but (R) is false.
Both (A) and (R) are true and (R) is the correct explanation of (A).

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(2)

Stopping potential, $V_{S} = \frac{h v - ϕ}{e}$ so stopping potential doesn't depend on intensity, stopping potential depends on frequency.

Intensity $\propto$ number of photons.

On increasing intensity, no. of photons per sec increases and so the no. of electrons.

Q 23 :

From the photoelectric effect experiment, following observations are made. Identify “which of these are correct.” [2023]

A. The stopping potential depends only on the work function of the metal.
B. The saturation current increases as the intensity of incident light increases.
C. The maximum kinetic energy of a photoelectron depends on the intensity of the incident light.
D. Photoelectric effect can be explained using wave theory of light.

Choose the correct answer from the options given below:

A, B, D only
B only
A, C, D only
B, C only

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(2)

(A) Stopping potential depends on both frequency of light and work function.

(B) Saturation current $\propto$ intensity of light

(C) Maximum KE depends on frequency

(D) Photoelectric effect is explained using particle theory.

Q 24 :

Given below are two statements: [2023]

Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.

Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.

In the light of above statements, choose the most appropriate answer from the options given below.

Both Statement I and Statement II are incorrect
Statement I is correct but statement II is incorrect
Both statement I and statement II are correct
Statement I is incorrect but statement II is correct

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(3)

Stopping potential $V_{s} = \frac{K E_{\max}}{e}$

$V_{s} = \frac{\frac{h C}{λ} - ϕ}{e}$

Stopping potential does not depend on intensity or power of light used; it only depends on the frequency or wavelength of the incident light.

So both statements I and II are correct.

Q 25 :

The threshold wavelength for photoelectric emission from a material is 5500 $Å$ . Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a ______ [2023]

A. 75 W infra-red lamp
B. 10 W infra-red lamp
C. 75 W ultra-violet lamp
D. 10 W ultra-violet lamp

Choose the correct answer from the options given below:

B and C only
A and D only
C only
C and D only

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(4)

$λ < 5500 Å for photoelectric emission$

$λ_{u v} < 5500 Å$

Q 26 :

If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8 eV and 2.2 eV. Then choose the correct option. [2023]

Metal B will not emit photo-electrons
Both metals A and B will emit photo-electrons
Both metals A and B will not emit photo-electrons
Metal A will not emit photo-electrons

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(4)

$ϕ = \frac{h c}{λ} = \frac{1240}{350} eV = 3.54 eV$

$∴$ $Only metal B will emit photoelectron.$

Q 27 :

The threshold frequency of metal is $f_{0}$ . When the light of frequency $2 f_{0}$ is incident on the metal plate, the maximum velocity of photoelectron is $v_{1}$ . When the frequency of incident radiation is increased to $5 f_{0}$ , the maximum velocity of photoelectrons emitted is $v_{2}$ . The ratio of $v_{1}$ to $v_{2}$ is [2023]

$\frac{v_{1}}{v_{2}} = \frac{1}{2}$
$\frac{v_{1}}{v_{2}} = \frac{1}{8}$
$\frac{v_{1}}{v_{2}} = \frac{1}{16}$
$\frac{v_{1}}{v_{2}} = \frac{1}{4}$

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(1)

$K_{\max} = h f - h f_{0}$

For $f = 2 f_{0}$

$\frac{1}{2} m v_{1}^{2} = 2 h f_{0} - h f_{0} = h f_{0} \dots (i)$

For $f = 5 f_{0}$

$\frac{1}{2} m v_{2}^{2} = 5 h f_{0} - h f_{0} = 4 h f_{0} \dots (i i)$

From (i) and (ii),

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{2}$

Q 28 :

The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is [2023]

1.24
2
1
1.5

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(3)

$e V_{s} = k_{\max}$

$V_{s} = {\frac{h}{e}} f + {\frac{- ϕ}{e}}$

$Slope is independent of nature of metal$

$slope {(V_{s})}^{Gold} = slope {(V_{s})}^{Aluminium}$

Q 29 :

In photoelectric effect [2023]

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:

A and E only
A and C only
A and B only
B and C only

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(2)

$Intensity of light \propto number of photons \propto no. of photoelectrons \propto photocurrent$

$So, A is correct$

$K E_{\max} = h ν - ϕ$

$K E_{\max} depends on frequency$

$So, C is correct$

$So, A and C are correct$

Q 30 :

The variation of stopping potential $(V_{0})$ as a function of the frequency $(ν)$ of the incident light for a metal is shown in figure. The work function of the surface is [2023]

18.6 eV
2.98 eV
2.07 eV
1.36 eV

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(3)

$e V_{0} = h ν - ϕ$

$0 = h ν - ϕ$

$ϕ = h ν = 6.6 \times 10^{- 34} \times 5 \times 10^{14} = 33 \times 10^{- 20} J$

$ϕ = \frac{33 \times 10^{- 20}}{1.6 \times 10^{- 19}} = 2.07 eV$

Topic Question Set

Q 21 :

The work function of metal is 3 eV. The colour of the visible light that is required to cause emission of photoelectrons is [2025]

Q 26 :

If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8 eV and 2.2 eV. Then choose the correct option. [2023]

Q 28 :

The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is [2023]

Q 30 :

The variation of stopping potential $(V_{0})$ as a function of the frequency $(ν)$ of the incident light for a metal is shown in figure. The work function of the surface is [2023]

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