Q145 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
In case of photoelectric emission from certain metal, the cutoff frequency is $\nu$. If the radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant)
- A$\sqrt{\dfrac{h\nu}{2m}}$
- B$\sqrt{\dfrac{h\nu}{m}}$
- Correct answerC$2\left(\sqrt{\dfrac{h\nu}{m}}\right)$
- D$\sqrt{\dfrac{2h\nu}{m}}$
Solution
According to Einstein's photoelectric equation:
$K_{\text{max}} = h\nu_{\text{incident}} - h\nu_{\text{cutoff}}$
Given $\nu_{\text{incident}} = 3\nu$ and $\nu_{\text{cutoff}} = \nu$:
$\dfrac{1}{2}mv_{\text{max}}^2 = h(3\nu) - h\nu$
$\dfrac{1}{2}mv_{\text{max}}^2 = 2h\nu$
$v_{\text{max}}^2 = \dfrac{4h\nu}{m}$
$v_{\text{max}} = 2\sqrt{\dfrac{h\nu}{m}}$
Answer: $2\left(\sqrt{\dfrac{h\nu}{m}}\right)$
$K_{\text{max}} = h\nu_{\text{incident}} - h\nu_{\text{cutoff}}$
Given $\nu_{\text{incident}} = 3\nu$ and $\nu_{\text{cutoff}} = \nu$:
$\dfrac{1}{2}mv_{\text{max}}^2 = h(3\nu) - h\nu$
$\dfrac{1}{2}mv_{\text{max}}^2 = 2h\nu$
$v_{\text{max}}^2 = \dfrac{4h\nu}{m}$
$v_{\text{max}} = 2\sqrt{\dfrac{h\nu}{m}}$
Answer: $2\left(\sqrt{\dfrac{h\nu}{m}}\right)$
Hence, the correct answer is option: 'C'.