Physics | VITEEE 2018 | Discussion Page

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

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

$2V_{0}$

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

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

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

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Answer:

Explanation: