In experiment of photoelectric effect, the stopping potential for a given metal is ${ }^{\prime}…

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. $\mathrm{V}_{0}+\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}$
  2. $\mathrm{~V}_{0}-\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}$
  3. $\frac{V_{0}}{2}$
  4. $2 \mathrm{~V}_{0}$

Solution

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

Asked in: MHT CET 2020 (15 Oct Shift 1)

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