Question

An electron of hydrogen atom is revolving in third Bohr's orbit $(n=3)$. The number of revolutions it will undergo before making a transition to the second orbit $(n=2)$ is
(Assume the average life time of an excited state of the hydrogen atom is in the order of ${10}^{-8}s$ and Bohr radius $=5.3\times {10}^{-12}m$)

• A. $2.5\times {10}^6$ rev
• B. $3.5\times {10}^6$ rev
• C. $4.5\times {10}^6$ rev
• D. $1.5\times {10}^6$ rev

Answer 1

The correct option is A $2.5\times {10}^6$ rev

Angular lifetime of exited state $={10}^{-8}$
Radius of 3rd qrbit $=9a0$ [where is boheis radius]
$speed=\frac{2.18\times {10}^6}{3}m/s$
$frequency=\frac{2\pi \left(9a0\right)}{speed}$
So, Numbers of revolutions $=\frac{{10}^{-8}\times speed}{2\pi \left(9a0\right)}$
$=\frac{{10}^8\times 2.18\times {10}^6\times {10}^{11}}{2\times 3.14\times 9\times 5.3}$ revolutions
$\approx 2.5\times {10}^{6}revolutions.$