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# When a metallic surface is illuminated with monochromatic light of wavelength $\;\lambda\;$ and stopping potential for photo electron current is $\;3V_{0}\;$ . When some metallic surface is illuminated with a light of wavelength $\;2 \lambda\;$ , The stopping potentials is $\;V_{0}\;$ . The threshold wavelength for surface is

$(a)\;6 \lambda\qquad(b)\;4 \lambda\qquad(c)\;\large\frac{4 \lambda}{3}\qquad(d)\;8 \lambda$

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## 1 Answer

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Answer : (b) $\;4 \lambda$
Explanation :
Einstein's photoelectric equation :
$\large\frac{hc}{\lambda}=e\;(3V_{0})+W----(1)$
Here W=work function ; and $\;\large\frac{hc}{\lambda}=e(V_{0})+W----(2)$
Solving equations
(1)-(2)
$\large\frac{hc}{\lambda}=3\;\large\frac{hc}{2\lambda}-W-3W$
$2W=\large\frac{hc}{2 \lambda}$
$\large\frac{hc}{\lambda_{0}}=\large\frac{hc}{4 \pi}$
$\lambda_{0}=4 \lambda \qquad \; when\;\lambda_{0}=$ threshold wavelength
answered Feb 25, 2014 by

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