A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by

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1.05 $\times$ 10−34 J s Let after absorption of energy, the hydrogen atom goes to the nth excited state. Therefore, the energy absorbed can be written as `10.2 = 13.6 xx ($\frac{1}{1}$^2 - 1/n^2) $\Rightarrow$ 10.$\frac{2}{13}$.6 = 1- 1/n^2 rArr = 1/(n^2) = (13.6 - 10.2)/13.6 rArr = 1/n^2 = 3.$\frac{4}{13}$.6 rArr n^2 = 4 rArr n = 2` The orbital angular momentum of the electron in the nth state is given by L_n =(nh)/(2pi) Change in the angular momentum, `DeltaL = (2h)/(2pi) - h/(2pi) = h/(2pi) DeltaL = 1.05xx10^-34` Js

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