When the light of frequency \( 2v_0 \) (where \( v_0 \) is the threshold frequency) is incident on a photosensitive material, the maximum velocity of electrons emitted is \( v_1 \). When the frequency of the incident radiation is increased to \( 3v_{0} \), the maximum velocity of emitted electrons from the same material is \( v_2 \). The ratio of \( v_1 \) to \( v_2 \) will then be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1 : \sqrt{2} \)
By Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is the photon energy minus the work function \( h v_0 \).
\( \displaystyle \frac{1}{2} m v_{\max}^{2} = h v - h v_0 \)
For frequency \( 2v_0 \):
\( \displaystyle \frac{1}{2} m v_1^{2} = 2hv_0 - hv_0 = hv_0 \)
For frequency \( 3v_0 \):
\( \displaystyle \frac{1}{2} m v_2^{2} = 3hv_0 - hv_0 = 2hv_0 \)
Dividing, and taking the square root because the kinetic energy goes as the square of the speed:
\( \displaystyle \frac{v_1^{2}}{v_2^{2}} = \frac{1}{2} \;\Rightarrow\; \frac{v_1}{v_2} = \frac{1}{\sqrt{2}} \)