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Question

(C) The work function $\phi$ of the metal is given by $\phi = h
u$,where $
u$ is the cutoff (threshold) frequency.According to Einstein's photoelect

Vedclass · Accepted Answer

$\sqrt{\frac{2h
u}{m}}$

Vedclass · Answer

(C) The work function $\phi$ of the metal is given by $\phi = h
u$,where $
u$ is the cutoff (threshold) frequency.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by $K_{\max} = E - \phi$.Here,the incident frequency is $2
u$,so the incident energy $E = h(2
u) = 2h
u$.Substituting the values,we get $K_{\max} = 2h
u - h
u = h
u$.Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have $\frac{1}{2}mv_{\max}^2 = h
u$.Solving for $v_{\max}$,we get $v_{\max}^2 = \frac{2h
u}{m}$,which implies $v_{\max} = \sqrt{\frac{2h
u}{m}}$.

For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).

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