Question

The dissociation energy of $H_2$ is $430.53kJ$ $mo{l}^{-1}$. If $H_2$ is exposed to radiant energy of wavelength $253.7$ nm. What $\%$ of radiant energy will be converted into kinetic energy?

• A. $91.3\%$
• B. $8.7\%$
• C. $9.5\%$
• D. $90.5\%$

Answer 1

The correct option is B $8.7\%$

$E_{H-H}$ bond dissociation$=\frac{430.53\times {10}^3}{6.023\times {10}^{23}}J$ per molecule

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

$E_{photon}=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{253.7\times {10}^{-9}}$

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

Energy converted into kinetic energy $=$ Energy left after dissociation of the bond.

$\therefore$ Energy converted into $KE=(7.83-7.15)\times {10}^{-19}J$

$=0.68\times {10}^{-19}J$

$\therefore \%$ of energy converted into $KE=\frac{0.68\times {10}^{-19}}{7.83\times {10}^{-19}}\times 100$

$=8.68\%$

Hence, the correct option is $B$