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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E. = h n - \phi$,where $\phi$ i

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

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K.E. = h n - \phi$,where $\phi$ is the work function of the metal.For the first photocathode: $\frac{1}{2}mV_1^2 = hn_1 - \phi$  $(1)$For the second photocathode: $\frac{1}{2}mV_2^2 = hn_2 - \phi$  $(2)$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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