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.
100%
B.
8.82%
C.
2.22%
D.
1.22%
Answer :
8.82%
Solution :
Energy of $$1\,mole$$ of photons,
$$\eqalign{
& E = {N_0} \times h\,\upsilon \cr
& \,\,\,\,\,\, = \frac{{{N_0} \times h \times c}}{\lambda } \cr
& \,\,\,\,\,\, = \frac{{6.023 \times {{10}^{23}} \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{253.7 \times {{10}^{ - 9}}}} \cr} $$
Energy converted into $$KE = \left( {472.2 - 430.53} \right)kJ$$
$$\% $$ of energy converted into
$$\eqalign{
& KE = \frac{{\left( {472.2 - 430.53} \right)}}{{472.2}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\, = 8.82\% \cr} $$