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

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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

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