CBSE-XI-Physics
43: Bohr's Model and Physics of the Atom
- #11A 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 sdigAnsr: aAns : (a) 1.05 × 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\times \left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)``
`` \Rightarrow \frac{10.2}{13.6}=1-\frac{1}{{n}^{2}}``
`` \Rightarrow \frac{1}{{n}^{2}}=\frac{13.6-10.2}{13.6}``
`` \Rightarrow \frac{1}{{n}^{2}}=\frac{3.4}{13.6}``
`` \Rightarrow {n}^{2}=4``
`` \Rightarrow n=2``
`` ``
The orbital angular momentum of the electron in the nth state is given by
`` {L}_{\,\mathrm{\,n\,}}=\frac{nh}{2\pi }``
Change in the angular momentum, `` ∆L=\frac{2h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }``
∴ `` ∆L=1.05\times {10}^{-34}\,\mathrm{\,Js\,}``
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