Question

The dissociation energy of $H_2$ is 430.53 kJ $mo{l}^{-1}$. If hydrogen is dissociated by illumination with radiation of wavelength 253.7 nm, the fraction of the radiant energy which will be converted into kinetic energy is given by

• A. 1.22%
• B. 8.76%
• C. 2.22%
• D. 100%

Answer 1

The correct option is B

8.76%

Energy of 1 mol of quanta $=hv{N}_A=\frac{hc{N}_4}{\lambda }$
$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times 6.23\times {10}^{23}}{253.7\times {10}^{-9}}=472.2kJ$
Energy converted into kinetic energy
= 472.2 - 430.53 kJ
% of radiant energy converted into kinetic energy
$=\frac{(472.2-430.53)\times 100}{472.2}=8.82$