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JEE Main 31-Jan-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 31-Jan-2024 Paper - Shift 14 Marks
MCQ
When a metal surface is illuminated by light of wavelength $5 \lambda$, the stopping potential is $8V.$ When the same surface is illuminated by light of wavelength $3 \lambda$, stopping potential is $2V.$ The threshold wavelength for this surface is $:$
  • A
    $5 \lambda$
  • B
    $3 \lambda$
  • C
    $9 \lambda$
  • D
    $4.5 \lambda$
✓

Answer

$E=\phi+K_{\max }$
$\phi=\frac{hc}{\lambda_0}$
$K_{\max }=eV_0$
$8 e=\frac{hc}{\lambda}-\frac{hc}{\lambda_0} \ldots \ldots .(i)$
$2 e=\frac{hc}{3 \lambda}-\frac{hc}{\lambda_0} \ldots \ldots . \text { (ii) }$
on solving $(i)\ \& \ (ii) $
$\lambda_0=9 \lambda$

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