The stopping potential doubles when the frequ | vedclass

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Question

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi$,where $\phi$ is the work function.For

Vedclass · Accepted Answer

$\frac{hv}{2}$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h
u - \phi$,where $\phi$ is the work function.For frequency $
u$,the stopping potential is $V_s = \frac{h
u - \phi}{e}$.For frequency $\frac{3
u}{2}$,the stopping potential is $2V_s = \frac{h(\frac{3
u}{2}) - \phi}{e}$.Substituting the first equation into the second: $2(\frac{h
u - \phi}{e}) = \frac{1.5h
u - \phi}{e}$.Multiplying by $e$: $2h
u - 2\phi = 1.5h
u - \phi$.Rearranging the terms: $2h
u - 1.5h
u = 2\phi - \phi$.Therefore,$\phi = 0.5h
u = \frac{h
u}{2}$.

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be

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