The dissociation energy of $H$ , is 430.53 KJ/mol. If $H_{2}$ is exposed to radiant energy of wavelength 253.7 nm, what percentage of radiant energy will be converted into kinetic energy?

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Solution

Dissociation energy of H per atom $= 450.53 \times \frac{1}{6.023 \times 10^{23}} = 7.48 \times 10^{- 22} k J = 7.48 \times 10^{- 19} J = 4.675 e V$
$E_{r a d i a t i o n} = \frac{h c}{λ} = \frac{1242}{253.7} = 4.9 e V E_{r a d i a t i o n} = E_{d i s s o c i a t i o n} + K E K E = 4.9 - 4.675 = 0.225 e V$
% Conversion of $E_{r a d i a t i o n}$ into $K E = \frac{0.225}{4.9} \times 100 = 4.6$ %

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

Dissociation energy of H per atom=450.53×16.023×1023=7.48×10−22kJ=7.48×10−19J=4.675eVEradiation=hcλ=1242253.7=4.9eVEradiation=Edissociation+KEKE=4.9−4.675=0.225eV% Conversion of Eradiation into KE=0.2254.9×100=4.6%

The dissociation energy of $H$ , is 430.53 KJ/mol. If $H_{2}$ is exposed to radiant energy of wavelength 253.7 nm, what percentage of radiant energy will be converted into kinetic energy?