Question

Average lifetime of a hydrogen atom excited to n=2 state is ${10}^{-8}$s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.

Answer 1

Velocity of the revolution of an electron is given by

$v_n=\frac{Z{e}^2}{2nh{\epsilon }_0}$

where

$z=1$

$n=1$

$h=6.626\times {10}^{-34}joulesec$

${\epsilon }_0=8.85\times {10}^{-12}farad/meter$

$e=1.6\times {10}^{-19}$

Thus,

$v_1=\frac{(1.6\times {10}^{-19}{)}^2}{2\times (6.626\times {10}^{-34})\times (8.85\times {10}^{-12})}$

$v_1=2.18\times {10}^{6}m/s$

$v_2=\frac{v_1}{2}=1.09\times {10}^{6}m/s$

$r_2=4r_1=4\times 0.0529\times {10}^{-9}m$

$\omega =\frac{2\pi }{T}=\frac{v_2}{r_2}$

$T=\frac{2\pi r_2}{v_2}$

$=\frac{2\pi \times 0.0529\times {10}^{-9}\times 4}{1.094\times {10}^6}$

$=1.22\times {10}^{-15}sec$

In $1.22\times {10}^{-15}sec$ electron makes one revolution.

In ${10}^{-8}sec$, electron will make$=\frac{{10}^{-18}}{1.22\times {10}^{-15}}=8.2\times {10}^{6}revolutions$