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 electron drops from this level to the next level. Find the wavelength of this line?

• A. $18.75\times {10}^{-7}$ m
• B. $1.87\times {10}^{-7}$ m
• C. $187.5\times {10}^{-7}$ m
• D. $0.188\times {10}^{-7}$ m

Answer 1

The correct option is D $0.188\times {10}^{-7}$ m

According to Bohr's atomic model
mvr = nh/2 $\pi$
$4.2176\times {10}^{-34}=n_2\times \frac{6.62\times {10}^{-34}}{2\times 3.14}$
$n_2=4$
So, if $n_2$ = 4, electrons is deexcited, so $n_1$ = 3
$\frac{1}{\lambda }=R{Z}^2\times [\frac{1}{3^2}-\frac{1}{4^2}]$
where Z = Atomic number
On solving, $\lambda =18.787\times {10}^{-7}$ m