Question

The angular momentum of an electron in a Bohr's orbit of H-atom is $4.2178\times {10}^{-34}{kgm}^2/sec$. Calculate the wavelength of the spectral line emitted when electrons falls from this level to next lower level.

Answer 1

$L=4.2178\times {10}^{-34}$

$\Rightarrow \frac{n\times h}{2\pi }=4.2178\times {10}^{-34}$

$\Rightarrow n=\frac{4.2178\times {10}^{-34}\times 2\pi }{6.626\times {10}^{-34}}$

$=\frac{4.2178\times 2\pi }{6.625}$

$=4$

$\therefore$ we have to calculate $\lambda$ when $4\to 3$

$\frac{1}{\lambda }=R_H\times z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\Rightarrow \frac{1}{\lambda }=R_H\times 1(\frac{1}{9}-\frac{1}{16})$

$\Rightarrow \lambda =\frac{144}{7R_H}$