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When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8 \mathrm{~V}$. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6 \mathrm{~V}$. The threshold wavelength of the metal is :
Correct Option: , 2
$\mathrm{V}_{\mathrm{S}}=\mathrm{h} v-\phi$
$4.8=\frac{\mathrm{hc}}{\lambda}-\phi$..(1)
$1.6=\frac{\mathrm{hc}}{2 \lambda}-\phi$..(2)
Using above equation (i) - (ii)
$3.2=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{2 \lambda}$
$3.2=\frac{\mathrm{hc}}{2 \lambda}$..(3)
$\left[\lambda=\frac{\mathrm{hc}}{6.4}\right]$
Put in equation (ii)
$\phi=1.6$
$\frac{\mathrm{hc}}{\lambda_{\mathrm{th}}}=1.6$
$\lambda_{\text {th }}=\frac{\text { hc }}{1.6}$
$=\left(\frac{\mathrm{hc}}{6.4}\right) \times 4=4 \lambda$
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