Modern Physics Ques 149
- Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65$ $ \AA$ ). The de-Broglie wavelength of this electron is
(Main 2019, 12 April II)
(a) $3.5 $ $\AA$
(b) $6.6 $ $\AA$
(c) $12.9 $ $\AA$
(d) $9.7 $ $\AA$
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Answer:
Correct Answer: 149.(d)
Solution:
- By Bohr’s IInd postulate, for revolving electron,
$ \begin{gathered} \quad \text { Angular momentum }=\frac{n h}{2 \pi} \Rightarrow m v r _n=\frac{n h}{2 \pi} \\ \Rightarrow \text { Momentum of electron, } p=m v=\frac{n h}{2 \pi r _n} \end{gathered} $
de-Broglie wavelength associated with electron is
$\lambda _n=\frac{h}{p}=\frac{2 \pi r _n}{n} $
$\text { Given, } n =3, r _n=4.65 \AA $
$\therefore \quad \lambda _n =\frac{(2 \times \pi \times 4.65)}{3} \approx 9.7 \AA$
BETA
