Question

The velocity of electron in a certain Bohr orbit of $H$-atom bears the ratio $1:275$ to the velocity of light.
Calculate the wave numbers of radiation when electrons jumps from $(n+1)$ state to ground state. Express your answer in p, where wave number = $p\times {10}^{4}c{m}^{-1}$

• A. 9.75
• B. 3.2
• C. 83
• D. 4

Answer 1

The correct option is A 9.75

Given velocity of electron in a certain Bohr orbit of $H$-atom $=\left(1/275\right)\times$ velocity of light
$=\left(1/275\right)\times 3\times {10}^{10}=1.09\times {10}^{8}cmse{c}^{-1}$
(a) Since, $u_n=\frac{2\pi e^2}{nh}$
$\therefore 1.09\times {10}^8=\frac{2\times 3.14\times (4.803\times {10}^{-10}{)}^2}{6.626\times {10}^{-27}\times n}$
$\therefore n=2.006=2$(an integer value)
(b) Thus, during the jump of electron from $(n+1)$,i.e., $3rd$ shell to ground state
$\overline{v}=\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{3^2}]=109678[\frac{1}{1}-\frac{1}{9}]$
$=9.75\times {10}^{4}c{m}^{-1}$