Discussion Forum



1)

In a photoelectric effect measurement, the stopping potential for a given metal is found to be V0 volt when radiation of wavelength λ0 is used.lf radiation of wavelength 2λ0 is used with the same metal then the stopping potential (in volt) will be


A) $\frac{V_{0}}{2}$

B) $2V_{0}$

C) $V_{0}+\frac{hc}{2e\lambda_{0}}$

D) $V_{0}-\frac{hc}{2e\lambda_{0}}$

Answer:

Option D

Explanation:

$eV_{0} =\frac{hc}{\lambda_{0}}-W_{0}$
and $eV' = \frac{hc}{2\lambda_{0}}-W_{0}$

Subtracting them we have

$e=\left(V_{0}-V'\right) =\frac{hc}{\lambda_{0}}\left[1-\frac{1}{2}\right] =\frac{hc}{2\lambda_{0}}$

or $V' =V_{0}-\frac{hc}{2e\lambda_{0}}$