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

Explanation:

The electron’s frequency (f) is shown as $f=\frac{m e^{4}}{4} \in 02 n 3 h 3$

Time period is shown as $T=\frac{1}{f}$

$T=\frac{4 \in_{0}^{2} n^{3} h^{3}}{m} e$

where,

h = Constant of Planck

m = electron’s mass

e = electron’s charge  

ε0 = Free space’s permittivity

Therefore, $T=4 \times(8.85) 2 \times 23 \times(6.63) 3 \times 10^{-24} \times 10-\frac{102}{9.1} \times(1.6) 4 \times 10^{-76}$

$=12247.735 \times 10^{-19} \mathrm{s}$

Hydrogen has the average lifetime $t=10^{-8} \mathrm{s}$

The revolutions number is shown as N = tT

$\Rightarrow N=8.2 \times 10^{5} \text { revolutions. }$