• # question_answer If the maximum kinetic energy, of emitted photoelectron from a metal surface of work function $2.5\text{ }eV$ is$1.7\text{ }eV$. If wavelength of incident radiation is halved, then stopping potential will be: A)  $2.5\,V$                    B)  $5.9\,V$C)  $5\,V$                       D)  $1.1\,V$

Solution :

As energy associated with the photon is given by: $\frac{hc}{\lambda }=h\upsilon$ $=2.5+1.7=4.2eV$ and           $\upsilon \lambda =c$ Now,         $V'\frac{\lambda }{2}=c$ $\Rightarrow$               $V'=2V$ $\Rightarrow$               $e{{V}_{0}}=h\,(2V)-\phi$                 $=2\times 4.2-2.5$                 ${{V}_{0}}=5.9V$

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