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MHT CET · Physics · Dual Nature of Matter

In experiment of photoelectric effect, the stopping potential for a given metal is
\({ }^{\prime} \mathrm{V}_{0}{ }^{\prime}\) volt, when radiation of wavelength \({ }^{\prime} \lambda_{0}{ }^{\prime}\) is used. If radiation of wavelength \({ }^{\prime} 2 \lambda_{0}{ }^{\prime}\)
is used for the same metal, then the stopping potential (in volt) will be \([\mathrm{e}=\) charge
on electron, \(\mathrm{c}=\) speed of light, \(\mathrm{h}=\) Planck's constant.]

  1. A \(\mathrm{V}_{0}+\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}\)
  2. B \(\mathrm{~V}_{0}-\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}\)
  3. C \(\frac{V_{0}}{2}\)
  4. D \(2 \mathrm{~V}_{0}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{~V}_{0}-\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}\)

Step-by-step Solution

Detailed explanation

\(h v-\omega_{0}=e V_{0}\)
\(\frac{h c}{\lambda_{0}}-\omega_{0}=e V_{0}\)
\(\frac{h c}{2 \lambda_{0}}-\omega_{0}=e V_{s}\)
\(=\frac{h c}{\lambda_{0}}\left[1-\frac{1}{2}\right]=e\left(V_{0}-V_{s}\right)=e V_{0}-e V_{s}\)
\(V_{s}=V_{0}-\frac{h c}{2 \lambda_{0} e}\)