Question

A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
(a) 1.05 × 10−34 J s
(b) 2.11 × 10−34 J s
(c) 3.16 × 10−34 J s
(d) 4.22 × 10−34 J s

Answer 1

Answer:

write properly question e

Answer 2

The orbital angular momentum of the body increases when it absorbs an energy of 10.2 eV is $1.05 \times 10^{-34} J s$

Explanation:

We know that according to Bohr’s model, The energy ∆E released by the electron when it moves from one energy level n1 to other energy level n2 is given by,

$\Delta E=13.6\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

As the given molecule is hydrogen Atom, Z can be considered as 1 and by considering the energy ∆E as 10.2 eV and n1 as 1 and n2 as n we get,

$1-\frac{1}{n^{2}}=\frac{10.2}{13.6} \quad \text { or } n=2$

Thus the increase in the angular moment is given by,

$L=\frac{h}{2 \pi}=\frac{6.625 X 10^{-34}}{2 \pi}=1.05 \times 10^{-34} \mathrm{Js}$

Hence option $1.05 \times 10^{-34} J s$ is the correct answer.