10.3 Orbital frequency of electron

  Basic Modern Physics

$$\begin{equation} \begin{aligned} f = \frac{v}{{2\pi r}} = \frac{{\left( {\frac{{Z{e^2}}}{{2{\varepsilon _0}nh}}} \right)}}{{2\pi \left( {\frac{{{n^2}{h^2}{\varepsilon _0}}}{{\pi m{e^2}Z}}} \right)}} = \frac{{m{Z^2}{e^4}}}{{4{\varepsilon _0}^2{n^3}{h^3}}} \\ \quad \Rightarrow f \propto \frac{{{Z^2}}}{{{n^3}}} \\\end{aligned} \end{equation} $$

Question 10. If the average life time of an excited state of hydrogen is of ${10^{ - 8}}\;s$, estimate how many orbits an electron makes when it is in the state $n=2$ and before it suffers a transition to state $n=1$ (Bohr radius ${a_0} = 5.3 \times {10^{ - 11}}\ m$).

Solution: The angular momentum of an electron in $nth$ orbit is $nh/2\pi $.

By Bohr's hypothesis: Again, $mvr = nh/2\pi $ where $r$ is the radius of the orbit.

$${\text{or }}\quad \quad v = \frac{{nh}}{{2\pi mr}}$$

The time period of completing an orbit is

$$\begin{equation} \begin{aligned} \quad \quad \quad T = \frac{{2\pi r}}{v} = \frac{{2\pi r}}{{\left( {\frac{{nh}}{{2\pi mr}}} \right)}} \\ {\text{or }}\quad \quad T = \frac{{4{\pi ^2}m{r^2}}}{{nh}} \\\end{aligned} \end{equation} $$

Since the radius of the orbit is proportional to ${n^2}$, hence

$$\begin{equation} \begin{aligned} \quad \quad \quad r = {a_0}{n^2} \\ \therefore \quad \quad T = \frac{{4{\pi ^2}ma_0^2{n^4}}}{{nh}} = \frac{{4{\pi ^2}ma_0^2{n^3}}}{h} \\ {\text{Number of orbits completed in 1}}{{\text{0}}^{ - 8}}\ s \\ \quad \quad = \frac{{{{10}^{ - 8}}}}{T} = \frac{{{{10}^{ - 8}} \times h}}{{4{\pi ^2}ma_0^2{n^4}}} \\ \quad \quad = \frac{{{{10}^{ - 8}} \times \left( {6.6 \times {{10}^{ - 34}}} \right)}}{{4{{\left( {3.14} \right)}^2}\left( {9.1 \times {{10}^{ - 31}}} \right){{\left( {5.3 \times {{10}^{ - 11}}} \right)}^2}{{\left( 2 \right)}^3}}} = 8 \times {10^6} \\\end{aligned} \end{equation} $$