Question

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

• A. $18.75\times {10}^{-7}m$
• B. $1.87\times {10}^{-7}m$
• C. $187.5\times {10}^{-7}m$
• D. $0.187\times {10}^{-7}m$

Answer 1

Answer: (A)

$18.75\times {10}^{-7}m$

According to Bohr's atomic model:

$mvr=\frac{nh}{2\pi }$

$4.2176\times {10}^{-34}=n_2\times \frac{6.62\times {10}^{-34}}{2\times 3.14}$

$\Rightarrow n_2=4$

When the electron drop,

$\frac{1}{\lambda }=R\times 1^2[\frac{1}{3^2}-\frac{1}{4^2}]$

$\lambda =18.75\times {10}^{-7}m$.