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(D) According to Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is given by $K_{max} = h
u - \phi$, where

Vedclass · Accepted Answer

$V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$

Vedclass · Answer

(D) According to Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is given by $K_{max} = h
u - \phi$, where $\phi$ is the work function of the metal.Since the photocathodes are identical, they have the same work function $\phi$.For frequency $n_1$, the kinetic energy is $\frac{1}{2}mV_1^2 = hn_1 - \phi$  --- $(1)$For frequency $n_2$, the kinetic energy is $\frac{1}{2}mV_2^2 = hn_2 - \phi$  --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$\frac{1}{2}mV_1^2 - \frac{1}{2}mV_2^2 = (hn_1 - \phi) - (hn_2 - \phi)$$\frac{1}{2}m(V_1^2 - V_2^2) = h(n_1 - n_2)$$V_1^2 - V_2^2 = \frac{2h}{m}(n_1 - n_2)$.

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)

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