WBJEE 2024 Physics & Chemistry Paper

To determine from which metals electrons will be emitted, we need to determine if the energy of the incident monochromatic light is greater than the work function of the metal. The energy of the incident light can be calculated using the formula:
E=‌

hc

λ

where:
h is Planck's constant: 6.626×10−34‌Js
c is the speed of light: 3×108m∕ s
λ is the wavelength of the light in meters
Given that the wavelength λ=4770

o

A

(which is equivalent to 4770×10−10m ), we can substitute into the formula:

E=‌

6.626×10−34×3×108

4770×10−10

By calculating this, we get:

E≈4.155×10−19J
To convert this energy from joules (J) to electron volts (eV), we use the conversion factor of
‌1‌eV=1.602×10−19J
‌E≈‌

4.155×10−19

1.602×10−19

‌eV
This simplifies to:

E≈2.59‌eV
Now, we compare this energy with the work functions of the metals A,B,C, and D :
A: 4.2 eV
B: 3.7 eV
C: 3.2 eV
D: 2.3 eV
Since the energy of the incident light (2.59‌eV) is less than the work functions of metals A(4.2‌eV), B(3.7‌eV), and C(3.2‌eV), electrons will not be emitted from these metals. However, the energy is greater than the work function of metal D(2.3‌eV), so electrons will be emitted from metal D .
Therefore, the correct answer is:
Option D: D only