$(a)\;5268A^{0}\qquad(b)\;3128A^{0}\qquad(c)\;1264A^{0}\qquad(d)\;4133A^{0}$

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Answer : (d) $\;4133 A^{0}$

Explanation :

$K_{1}=\large\frac{12400}{3000}-W=4.13-W----->(1)$

$K_{2}=\large\frac{12400}{1650}-W=7.51-W----->(2)$

Since $\;V_{2}=2V_{1} \quad \; S0\;,K_{2}=4K_{1}----->(3)$

Solving above equations .

$W=3eV$

Therefore , Threshold wavelength $\;\lambda_{0}=\large\frac{12400}{3}=4133A^{0}\;.$

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