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. `8.86%`
B. `2.33%`
C. `1.3%`
D. `9.56%`

Answer

A

Explanation

`(hc)/(lamda)=(430.53xx10^(3))/(6.023xx10^(23))+K.E.`
`K.E =(6.626xx10^(-34)xx3xx10^(8))/(253.7xx10^(-9))-(430.53xx10^(3))/(6.023xx10^(23))`
`=6.9xx10^(-20)`
`:.` Fraction `=(6.9xx10^(-20))/(7.86xx10^(-19))=0.088`
`=8.86%`
Hence a is the correct answer.