Q145 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass m are $V_1$ and $V_2$ respectively, then ($h$ = Planck's constant)
- A$V_1 + V_2 = \left[\dfrac{2h}{m}(n_1 + n_2)\right]^{\frac{1}{2}}$
- B$V_1 - V_2 = \left[\dfrac{2h}{m}(n_1 - n_2)\right]^{\frac{1}{2}}$
- C$V_1^2 + V_2^2 = \dfrac{2h}{m}(n_1 + n_2)$
- Correct answerD$V_1^2 - V_2^2 = \dfrac{2h}{m}(n_1 - n_2)$
Solution
Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K = hn - W$, where $W$ is the work function of the material.
For the first case:
$\dfrac{1}{2} m V_1^2 = h n_1 - W$
For the second case:
$\dfrac{1}{2} m V_2^2 = h n_2 - W$
Subtracting the second equation from the first, we get:
$\dfrac{1}{2} m V_1^2 - \dfrac{1}{2} m V_2^2 = h n_1 - W - (h n_2 - W)$
$\dfrac{1}{2} m (V_1^2 - V_2^2) = h(n_1 - n_2)$
$V_1^2 - V_2^2 = \dfrac{2h}{m}(n_1 - n_2)$
For the first case:
$\dfrac{1}{2} m V_1^2 = h n_1 - W$
For the second case:
$\dfrac{1}{2} m V_2^2 = h n_2 - W$
Subtracting the second equation from the first, we get:
$\dfrac{1}{2} m V_1^2 - \dfrac{1}{2} m V_2^2 = h n_1 - W - (h n_2 - W)$
$\dfrac{1}{2} m (V_1^2 - V_2^2) = h(n_1 - n_2)$
$V_1^2 - V_2^2 = \dfrac{2h}{m}(n_1 - n_2)$
Hence, the correct answer is option: 'D'.