Question

The angular momentum of an electron in a Bohr's orbit of $H e^{+}$ is $3.1652\times 10^{- 34}$ $kg \, m^{2} / s e c$ . What is the wave number in terms of Rydberg constant $\left(R\right)$ of the spectral line emitted when an electron falls from this level to the first excited state. $\left[Use \, h = 6.626 \times 10^{- 34} \, J . s\right]$

• A. $3R$
• B. $\frac{5 R}{9}$
• C. $\frac{3 R}{4}$
• D. $\frac{8 R}{9}$

Answer 1

Answer: (B)

Angular momentum $=\frac{ nh }{2 \pi}$
$3.1652 \times 10^{-34}=\frac{n \times 6.626 \times 10^{-34}}{2 \pi}$
$n =3$
$\because \bar{v}=R \cdot Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\overline{ v }= R \cdot 2^2 \frac{1}{2^2}-\frac{1}{3^2} \Rightarrow \frac{5 R }{9}$