Q114 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
The threshold frequency of metal is $f_0$. When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5f_0$ the maximum velocity of photoelectrons emitted is $v_2$. The ratio $v_1$ to $v_2$ is
- Correct answerA$1:2$
- B$1:8$
- C$1:16$
- D$1:4$
Solution
Using Einstein's photoelectric equation:
$K_{\max} = hf - hf_0$
For incident frequency $2f_0$:
$\dfrac{1}{2}mv_1^2 = h(2f_0) - hf_0 = hf_0$
For incident frequency $5f_0$:
$\dfrac{1}{2}mv_2^2 = h(5f_0) - hf_0 = 4hf_0$
Taking the ratio of the two equations:
$\dfrac{v_1^2}{v_2^2} = \dfrac{hf_0}{4hf_0} = \dfrac{1}{4}$
$\dfrac{v_1}{v_2} = \dfrac{1}{2}$
Answer: $1:2$
$K_{\max} = hf - hf_0$
For incident frequency $2f_0$:
$\dfrac{1}{2}mv_1^2 = h(2f_0) - hf_0 = hf_0$
For incident frequency $5f_0$:
$\dfrac{1}{2}mv_2^2 = h(5f_0) - hf_0 = 4hf_0$
Taking the ratio of the two equations:
$\dfrac{v_1^2}{v_2^2} = \dfrac{hf_0}{4hf_0} = \dfrac{1}{4}$
$\dfrac{v_1}{v_2} = \dfrac{1}{2}$
Answer: $1:2$
Hence, the correct answer is option: 'A'.