The dissociation energy of `H_(2)` is `430.53 kJ mol^(-1), `If `H_(2)` is of dissociated by illumination with radiation of wavelength `253.7 nm` , the fraction of the radiant energy which will be converted into ikinetic energy is given by

Options

A. `7.22%`
B. `8.82%`
C. `2.22%`
D. `100%`

Answer

B

Explanation

Energy of 1 mole of quanta = `hv N_(A) = (hc N_(A))/(lambda)`
` = (6.63 xx 10^(-34) xx 3 xx 10^(8) xx 6.023 xx 10^(23))/(253.7 xx 10^(-9))`
Energy converted into kinetic energy
` = 472.7 - 430.53 kJ`
% of radiant energy converted into kinetic energy
`= ((472.2 - 430.53)xx 100)/(472.2) = 8.82%`