Question

The angular momentum of an electron in a Bohr's orbit of H-atom is $3.1652\times {10}^{-34}kg.m^2/sec.$ The wavenumber in terms of Rydberg constant (R) of the spectral line emitted when an electron falls from this level to the ground state is: [Use h = $6.626\times {10}^{-34}Js$]

• A. $R\left(\frac{8}{9}\right)$
• B. $R\left(\frac{9}{8}\right)$
• C. $R\left(\frac{6}{9}\right)$
• D. $R\left(\frac{9}{6}\right)$

Answer 1

The correct option is A $R\left(\frac{8}{9}\right)$

Angular momentum is given by:

$mvr=n\frac{h}{2\pi }$

$n\frac{6.62\times {10}^{-34}}{2\pi }=3.1652\times {10}^{-34}kg.m^2/sec.$

$\Rightarrow n=3$

$\overline{\nu }=\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$n_1=1,n_2=3$

$\overline{\nu }=\frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{3^2}]$

$\Rightarrow \overline{\nu }=\frac{8R}{9}$