Question

The dissociation energy of $H_2$ is $482.5KJ/mol$. If $H_2$ is exposed to radiant energy of wavelength 124nm, what $\%$ of radiant energy will be converted into K.E?

$(1\frac{ev}{atom}5=96kJ/mol.)$

Answer 1

Energy required to break H-H bond

$=\frac{482.5\times {10}^3}{6.023\times {10}^{23}}$ J/Molecules

$=8\times {10}^{-19}$ J

Energy of photon used for this purpose $=hc/\lambda$

$=\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{124\times {10}^{-9}}$

$=16\times {10}^{-19}$ J

Energy left $=(16-8)\times {10}^{-19}$

Energy converted into K.E $=8\times {10}^{-19}J$

% of energy used in kinetic energy $=\frac{8\times {10}^{-19}}{16\times {10}^{-19}}\times 100$

$=50$%