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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.]
Options:
${ }^{\prime} \mathrm{V}_{0}{ }^{\prime}$ 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.]
Solution:
2230 Upvotes
Verified Answer
The correct answer is:
$\mathrm{~V}_{0}-\frac{\mathrm{hc}}{2 \mathrm{e} \lambda_{0}}$
$h v-\omega_{0}=e V_{0}$
$\frac{h c}{\lambda_{0}}-\omega_{0}=e V_{0}$
$\frac{h c}{2 \lambda_{0}}-\omega_{0}=e V_{s}$
$=\frac{h c}{\lambda_{0}}\left[1-\frac{1}{2}\right]=e\left(V_{0}-V_{s}\right)=e V_{0}-e V_{s}$
$V_{s}=V_{0}-\frac{h c}{2 \lambda_{0} e}$
$\frac{h c}{\lambda_{0}}-\omega_{0}=e V_{0}$
$\frac{h c}{2 \lambda_{0}}-\omega_{0}=e V_{s}$
$=\frac{h c}{\lambda_{0}}\left[1-\frac{1}{2}\right]=e\left(V_{0}-V_{s}\right)=e V_{0}-e V_{s}$
$V_{s}=V_{0}-\frac{h c}{2 \lambda_{0} e}$
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