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

Frequency of electron (f) is given by
$f=\frac{m{e}^4}{4{{\in }_0}^2{n}^3{h}^3}$
Time period is given by
$$\begin{gathered} T=\frac{1}{f} \\ T=\frac{4{\in }_0^2{n}^3{h}^3}{m{e}^4} \end{gathered}$$
Here,
h = Planck's constant
m = Mass of the electron
e = Charge on the electron
${\epsilon }_0$= Permittivity of free space
$$\begin{gathered} \therefore T=\frac{4\times \left(8.85\times {10}^{-12}\right)\times {\left(2\right)}^3\times {\left(6.63\times {10}^{-34}\right)}^3}{\left(9.10\times {10}^{-31}\right)\times {\left(1.6\times {10}^{-16}\right)}^4} \\ =12247.735\times {10}^{-19}\mathrm{s} \end{gathered}$$
Average life time of hydrogen, t = 10^$-$8 s
Number of revolutions is given by
$$\begin{gathered} N=\frac{t}{T} \\ \Rightarrow N=\frac{{10}^{-8}}{12247.735\times {10}^{-19}} \end{gathered}$$
N = 8.2 $\times$ 10⁵ revolution