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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - \Phi$,where $
u$ is the frequen

Vedclass · Accepted Answer

$\left( \frac{n_2 - n_1}{n_1 - 1} \right) K_0$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - \Phi$,where $
u$ is the frequency and $\Phi$ is the work function.Initially,$K_0 = h
u - \Phi$  --- $(1)$When the frequency is increased by a factor of $n_1$,the new frequency is $
u' = n_1
u$. The new maximum kinetic energy is $n_2K_0$.So,$n_2K_0 = h(n_1
u) - \Phi$  --- $(2)$From $(1)$,$h
u = K_0 + \Phi$. Substituting this into $(2)$:$n_2K_0 = n_1(K_0 + \Phi) - \Phi$$n_2K_0 = n_1K_0 + n_1\Phi - \Phi$$n_2K_0 - n_1K_0 = \Phi(n_1 - 1)$$K_0(n_2 - n_1) = \Phi(n_1 - 1)$$\Phi = \left( \frac{n_2 - n_1}{n_1 - 1} ight) K_0$Thus,the correct option is $A$.

In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.

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