JEE Main Dual Nature of Radiation and Matter PYQ

16. (JEE Main 2022 (Online) 29th June Evening Shift)

The electric field at a point associated with a light wave is given by

E = 200 [sin (6 $\times$ 10¹⁵)t + sin (9 $\times$ 10¹⁵)t] Vm^{= $-$ 1}

Given : h = 4.14 $\times$ 10^{$-$ 15} eVs

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

(A) 1.90 eV

(B) 3.27 eV

(C) 3.60 eV

(D) 3.42 eV

Correct answer is option D

Frequency of EM waves = $\frac{6}{2 π} \times 10^{15}$ and $\frac{9}{2 π} \times 10^{15}$

Energy of one photon of these waves

$= (4.14 \times 10^{- 15} \times \frac{6}{2 π} \times 10^{15})$ eV

and $(4.14 \times 10^{- 15} \times \frac{9}{2 π} \times 10^{15})$ eV

= 3.95 eV and 5.93 eV

$\Rightarrow$ Energy of maximum energetic electrons

= 5.93 $-$ 2.50 = 3.43 eV

17.(JEE Main 2022 (Online) 29th June Morning Shift)

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.

(A) Both A and R are correct and R is the correct explanation of A.

(B) Both A and R are correct but R is not the correct explanation of A.

(C) A is correct but R is not correct.

(D) A is not correct but R is correct.

Correct answer is option B

When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.

18.( JEE Main 2022 (Online) 28th June Evening Shift)

Let K₁ and K₂ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $λ$ ₁ and $λ$ ₂, respectively are incident on a metallic surface. If $λ$ ₁ = 3 $λ$ ₂ then :

(A) $K_{1} > \frac{K_{2}}{3}$

(B) $K_{1} < \frac{K_{2}}{3}$

(C) $K_{1} = \frac{K_{2}}{3}$

(D) $K_{2} = \frac{K_{1}}{3}$

Correct answer is option B

$K_{1} = \frac{h c}{λ_{1}} - ϕ = \frac{h c}{3 λ_{2}} - ϕ$ ..... (i)

and $K_{2} = \frac{h c}{λ_{2}} - ϕ$ ..... (ii)

from (i) and (ii) we can say

$3 K_{1} = K_{2} - 2 ϕ$

$K_{1} < \frac{K_{2}}{3}$

19.(JEE Main 2022 (Online) 26th June Evening Shift )

A metal surface is illuminated by a radiation of wavelength 4500Å . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90^o with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

(A) 1.36 eV

(B) 1.69 eV

(C) 2.78 eV

(D) 2.23 eV

$\frac{h c}{λ} - ϕ = K E$ ...... (i)

$R = \frac{m v}{B q} = \frac{\sqrt{2 m (K E)}}{B q}$ ...... (ii)

Putting the values,

$ϕ ≃ 1.36$ eV

20.(JEE Main 2022 (Online) 24th June Evening Shift )

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

(A) 1 : 1

(B) 2 : 1

(C) 4 : 1

(D) 1 : 4

Correct answer is option B

$3.8 = 0.6 + \frac{1}{2} m v_{1}^{2}$

$1.4 = 0.6 + \frac{1}{2} m v_{2}^{2}$

$\Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{3.2}{0.8} = \frac{4}{1}$

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