JEE Main Dual Nature of Radiation and Matter PYQ

11. (JEE Main 2023 (Online) 24th January Morning Shift )

From the photoelectric effect experiment, following observations are made. Identify which of these are correct.

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 photo electron 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) A, B, D only

(B) A, C, D only

(C) B, C only

(D) B only

Correct answer is option D

(A) From Einstein's equation

$K_{max} = e V_{s} = h v - ϕ$

Form the stopping potential $(V_{s})$ depends on $ϕ$ $& v$ .

(B) Saturation current is proportional to intensity, i.e., number of incident photons.

(C) $K_{max}$ only depends on nature of photon and $ϕ$ .

(D) Einstein used particle behaviour of photon to explain photon electric effect.

Only B is correct.

12. (JEE Main 2022 (Online) 29th July Morning Shift )

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength:

(A) $\frac{hc}{E λ - hc}$

(B) $\frac{hc λ}{E λ + hc}$

(C) $\frac{h λ}{E λ + hc}$

(D) $\frac{hc λ}{E λ - hc}$

Correct answer is option B

$k = \frac{h c}{λ} - ϕ = E$

and, $2 k = \frac{h c}{λ_{2}} - ϕ = 2 E$

$\Rightarrow \frac{h c}{λ} - E = \frac{h c}{λ_{2}} - 2 E$

$\Rightarrow \frac{h c}{λ_{2}} = \frac{h c}{λ} + E$

$\Rightarrow λ_{2} = \frac{h c λ}{h c + λ E}$

13. (JEE Main 2022 (Online) 28th July Evening Shift )

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

(A) 1 : 2

(B) 1 : 3

(C) 2 : 3

(D) 3 : 2

Correct answer is option C

$\frac{1}{2} m v_{1}^{2} = 5 ϕ - ϕ$

And, $\frac{1}{2} m v_{2}^{2} = 10 ϕ - ϕ$

$\Rightarrow {(\frac{v_{1}}{v_{2}})}^{2} = \frac{4}{9}$

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{2}{3}$

14. (JEE Main 2022 (Online) 27th July Evening Shift )

With reference to the observations in photo-electric effect, identify the correct statements from below :

(A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.

(B) The value of saturation current increases on moving the source of light away from the metal surface.

(C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light.

(D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.

(E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves.

Choose the correct answer from the options given below :

(A) (A) and (B) only

(B) (A) and (E) only

(C) (A) and (E) only

(D) (D) and (E) only

Correct answer is option B

$∵$ $\frac{1}{2} m v_{m}^{2} = h ν - ϕ$

$\Rightarrow v_{m}^{2}$ varies linearly with frequency.

And, threshold wavelength can be explained by particle nature of light.

15.(JEE Main 2022 (Online) 25th July Morning Shift )

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 workfunction of the metal is : (Take hc = 1230eV-nm ).

(A) 1.537 eV

(B) 2.46 eV

(C) 0.615 eV

(D) 1.23 eV

Correct answer is option C

$∵$ $K_{m} = \frac{h c}{λ} - ϕ$

$\Rightarrow K = \frac{1230}{800} - ϕ$

and, $2 K = \frac{1230}{500} - ϕ$

$\Rightarrow 2 \times \frac{1230}{800} - 2 ϕ = \frac{1230}{500} - ϕ$

$\Rightarrow ϕ = 0.615 e V$