Question

The ratio of velocity of an electron in a certain Bohr's orbit ($n^{th}$ orbit) of H-atom to the velocity of light is $1:550$ . The wavenumber of radiation emitted when the electron jumps from the (n+1) to ground state:
(Rydberg constant $\left(R_H\right)=109678c{m}^{-1}$

• A. $9.75\times {10}^{4}c{m}^{-1}$
• B. $1.03\times {10}^{5}c{m}^{-1}$
• C. $1.57\times {10}^{4}c{m}^{-1}$
• D. $2.75\times {10}^{5}c{m}^{-1}$

Answer 1

The correct option is A $9.75\times {10}^{4}c{m}^{-1}$

We know that the velocity of electron in $n^{th}$ orbit of hydrogen like atom:
$V_e=2.189\times {10}^6\times \frac{Z}{n}m{s}^{-1}$
for hydrogen Z = 1
So, according to the question,
$\frac{V_e}{V_{light}}=\frac{1}{275}=\frac{2.189\times {10}^6\times \frac{Z}{n}}{3\times {10}^8}$
$n=2$
So, $n+1=3$, transition from $n=3$ to $n=1$
$\frac{1}{\lambda }=1^2\times 109678(\frac{1}{n_1^2}-\frac{1}{n_2^2})c{m}^{-1}$
$\frac{1}{\lambda }=1^2\times 109678(\frac{1}{1^2}-\frac{1}{3^2})c{m}^{-1}$
$\frac{1}{\lambda }=9.75\times {10}^{4}c{m}^{-1}$