Question

The angular momentum of an electron in a Bohr's orbit of hydrogen atom is $4.218\times {10}^{-34}kg{m}^2{s}^{-1}$. Calculate the wavelength of the spectral line emitted, when electrons falls from this level to the next lower level.

• A. $1.876\times {10}^{-4}$ cm
• B. $1.876\times {10}^{-4}$ m
• C. $1.786\times {10}^{-4}$ cm
• D. $1.786\times {10}^{-4}$ m

Answer 1

The correct option is A $1.876\times {10}^{-4}$ cm

Angular momentum of an electron in a Bohr's orbit of H-atom.
$mvr=\frac{nh}{2\pi }$
$4.218\times {10}^{-34}=\frac{n\times 6.626\times {10}^{-34}}{2\times \frac{22}{7}}$
$\Rightarrow n=\frac{4.218\times {10}^{-34}\times 2\times 22}{6.626\times {10}^{-34}\times 7}=4$
Now,
$\stackrel{-}{\nu }=\frac{1}{\lambda }=109678(\frac{1}{n_1^2}-\frac{1}{n_2^2})c{m}^{-1}$
The spectral lines when electron falls from $4^{th}$ level to $3^{rd}$ level,
$\frac{1}{\lambda }=109678(\frac{1}{3^2}-\frac{1}{4^2})c{m}^{-1}$
$\Rightarrow \frac{1}{\lambda }=109678\left(\frac{16-9}{9\times 16}\right)c{m}^{-1}$
$\Rightarrow \frac{1}{\lambda }=109678\left(\frac{7}{9\times 16}\right)c{m}^{-1}$
$\lambda =\frac{9\times 16}{109678\times 7}=1.876\times {10}^{-4}cm$