# The dissociation energy of $H_2$ is 430.53 KJ$mol^{-1}$. If $H_2$ is exposed to radiation energy of wavelength 253.7nm, what % of radiant energy will be converted into kinetic energy ?

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

## 1 Answer

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)
answered Feb 1, 2014

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