$(a)\;9.3\%\qquad(b)\;8.68\%\qquad(c)\;7.36\%\qquad(d)\;9.24\% $

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Energy required to break H-H bond

$=\large\frac{430.53\times 10^3}{6.023\times 10^{23}}$ J/Molecules

$=7.15\times 10^{-19}$ J

Energy of photon used for this purpose = $\large\frac{hc}{\lambda}$

$=\large\frac{6.625\times10^{-34}\times 3.0\times 10^8}{253.7\times10^{-9}}$

$=7.83\times 10^{-19}$ J

Energy left = $(7.83-7.15)\times 10^{-19}$

Energy converted into K.E = $0.68\times 10^{-19}$ J

% of energy used in kinetic energy = $\large\frac{0.68\times 10^{-19}}{7.83\times 10^{-19}}\times 100$

=8.68%

Hence answer is (b)

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