$(a)\;759nm\qquad(b)\;857nm\qquad(c)\;957nm\qquad(d)\;657nm$

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$E_{\text{photon absorbed}}=E_1+E_2$(energy released)

$\large\frac{hC}{\lambda}=\frac{hC}{\lambda_1}+\frac{hC}{\lambda_2}$

$\Rightarrow\large\frac{1}{\lambda}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$

$\Rightarrow\large\frac{1}{300\;nm}=\frac{1}{496\;nm}+\frac{1}{\lambda_2}$

$\therefore\lambda_2 = \Large\frac{496\times300}{496-300}$$\;nm=759.183\;nm$

Hence (a) is the correct answer.

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