Q200 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
Light of frequency $4\nu_0$ is incident on the metal of threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectron is ( h = Planck's constant)
- Correct answerA$3h\nu_0$
- B$2h\nu_0$
- C$\dfrac{3}{2}h\nu_0$
- D$\dfrac{1}{2}h\nu_0$
Solution
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is given by
$K_{\text{max}} = h\nu - h\nu_0$
Given that the frequency of incident light is $\nu = 4\nu_0$ and the threshold frequency is $\nu_0$.
Substituting the values, we get
$K_{\text{max}} = h(4\nu_0) - h\nu_0$
$K_{\text{max}} = 3h\nu_0$
Answer: $3h\nu_0$
$K_{\text{max}} = h\nu - h\nu_0$
Given that the frequency of incident light is $\nu = 4\nu_0$ and the threshold frequency is $\nu_0$.
Substituting the values, we get
$K_{\text{max}} = h(4\nu_0) - h\nu_0$
$K_{\text{max}} = 3h\nu_0$
Answer: $3h\nu_0$
Hence, the correct answer is option: 'A'.