Question

An electron in a Bohr orbit of hydrogen atom with the quantum number $n_2$ has an angular momentum $4.2176\times {10}^{-34}kg{m}^2{s}^{-1}$. If the electron drops from this level to the next lower level, the wavelength of this line is

• A. $18nm$
• B. $187.6pm$
• C. $1876\mathring{A}$
• D. $1.876\times {10}^4\mathring{A}$

Answer 1

Answer: (D)

$1.876\times {10}^4\mathring{A}$

Angular momentum of an electron in nth orbit is: $L=\frac{nh}{2\pi }$
Given, $L_{n_2}=4.2176\times {10}^{-34}=\frac{n_2(6.625\times {10}^{-34})}{2\pi }\Rightarrow n_2=4$
The next lower line is $n=3$
Therefore, wavelength emitted is:
$\frac{1}{\lambda }=(1.097\times {10}^7)(\frac{1}{9}-\frac{1}{16})\Rightarrow \lambda =18.75\times {10}^{-7}m$