The dissociation energy of H2 is 430.53 KJ/mol. If H2 is exposed to radiant energy of wavelength 253.7 nm, what % of radiant energy will be converted into KE?

The bond dissociation energy for the H-H bond is the energy required to break the H₂ molecule (in its most stable geometry) into separate hydrogen atoms.

Radiant energy = 1242 eV-nm / 253.7 nm = 4.895 eV = 7.843x10^-19 J

1 mole of electrons is 6.022 x 10²³ electrons

so radiant energy = 472.28 KJ/mol

dissoc. energy     = 430.53 KJ/mol

KE = rad. enrgy - dissoc. energy

      = 472.28 - 430.53

      =41.75 KJ/mol

% of rad. enrgy to KE is = 41.75*100/472.28

                                           = 8.84 %

--

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch