MCQ
When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8 \mathrm{~V}$. When the same surface is illuminated by light of wavelength $3 \lambda$, stopping potential is $2 \mathrm{~V}$. The threshold wavelength for this surface is :
  • A
    $5 \lambda$
  • B
    $3 \lambda$
  • $9 \lambda$
  • D
    $4.5 \lambda$

Answer

Correct option: C.
$9 \lambda$
c
$\mathrm{E}=\phi+\mathrm{K}_{\max }$

$\phi=\frac{\mathrm{hc}}{\lambda_0}$

$\mathrm{~K}_{\max }=\mathrm{eV}_0$

$8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) }$

$2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) }$

$\text { on solving (i) & (ii) }$

$\lambda_0=9 \lambda$

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