Question

The angular momentum of an electron in a Bohr's orbit of H -atom is $4\times {10}^{-34}kg{m}^2/sec.$ The wavelenth of spectral line emitted when an electron falls from this level to next lower level is $x\times {10}^{-4}$ cm. Find the value of $x$.
(Take, Planck's constant $h=6\times {10}^{-34}Js,\pi =3$, Rydberg constant $R_H=1.08\times {10}^7{m}^{-1}$ )

• A. 9.1
• B. 1.9
• C. 8.2
• D. 2.8

Answer 1

The correct option is B 1.9

According to Bohr's law,
$\text{angular momentum}\left(mvr\right)=\frac{nh}{2\pi }$
According to the question,
$\frac{nh}{2\pi }=4\times {10}^{-34}$
$\Rightarrow \frac{n\times (6\times {10}^{-34})}{2\times 3}=4\times {10}^{-34}$
On solving
$n=4$

We know,
$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Here,
($n_2=4,n_1=3)$

$\therefore \frac{1}{\lambda }=1.08\times {10}^7(\frac{1}{3^2}-\frac{1}{4^2})$
$\Rightarrow \frac{1}{\lambda }=1.08\times {10}^7\left(\frac{7}{144}\right)$
On solving
$\lambda =1.905\times {10}^{-6}m$
$=1.905\times {10}^{-4}cm$

Hence, $x\approx 1.9$