Question

The velocity of an electron in certain Bohr orbit of $H$-atom bears the ratio $1:275$ to the velocity of light. (a) What is the quantum number '$n$' of the orbit? (b) Calculate the wave number of the radiation when the electron jumps from $(n+1)$ state to ground state.

Answer 1

$\overline{v}=9.75\times {10}^4{cm}^{-1}$
(a) $\frac{v}{c}=\frac{1}{275}$ or $v=\frac{3\times {10}^{10}}{275}=1.09\times {10}^{8}cm{sec}^{-1}$
$v=\frac{nh}{2\pi mr}=\frac{nh}{2\pi m\times 0.529\times {10}^{-8}\times n^2}$
or $n=\frac{h}{2\pi m\times 0.529\times {10}^{-8}\times v}$
$=\frac{6.625\times {10}^{-27}}{2\times 3.14\times 9.1\times {10}^{-28}\times 0.529\times {10}^{-8}\times 1.09\times {10}^8}$
$=2$
(b) Thus, $n+1=2+1=3$. The electron jumps from $3$rd orbit to $1$st orbit.