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When radiation of wavelength

Question:

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 :

  1. $2 \lambda$

  2. $4 \lambda$

  3. $8 \lambda$

  4. $6 \lambda$


Correct Option: , 2

Solution:

$\mathrm{V}_{\mathrm{S}}=\mathrm{h} v-\phi$

$4.8=\frac{\mathrm{hc}}{\lambda}-\phi$..(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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