$(a)\;3.62\times10^{-19}J\qquad(b)\;1.623\times 10^{-19}J\qquad(c)\;2.625\times 10^{-19}J\qquad(d)\;4.8\times 10^{-19}J$

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$E_{photon-absorbed}=\large\frac{6.625\times10^{-34}\times3.0\times 10^8}{300\times10{-9}}$

$E_{re-emitted}=\large\frac{6.625\times10^{-34}\times3.0\times 10^8}{496\times10^{-9}}$

$=4.0\times10^{-19}J$

$E_{absorbed} = E_{photon}+E_{photon re-emitted}$

$E_{11} = 6.625\times 10^{-19}-4.0\times 10^{-19}$

$=2.625\times10^{-19}$

Hence the answer is (c)

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