Question

The dissociation energy of $H_2$ is 430.53 KJ/mol. If $H_2$ is exposed to radiant energy of wavelength 253.7nm, what % of radiant energy will be converted into K.E?

• A. 8.86%
• B. 7.08%
• C. 2.30%
• D. 9.05%

Answer 1

The correct option is A 8.86%

The energy of one photon of wavelength 253.7 nm is $E=\frac{hc}{\lambda }=\frac{6.6256\times {10}^{-34}Js\times 2.9979\times {10}^{8}m/s}{253.7\times {10}^{-9}m}=7.83\times {10}^{-19}J=7.83\times {10}^{-22}kJ$
The energy for one mole of photons is $E=6.023\times {10}^{23}\times 7.83\times {10}^{-22}kJ=472kJ/mol.$
Out of this, the radiant energy converted into kinetic energy is $472kJ/mol-430.53kJ/mol=41.47kJ/mol$
The percentage of the radiant energy converted into kinetic energy is $\frac{41.47kJ/mol}{472kJ/mol}\times 100=8.86$%