An electron in a hydrogen atom makes a transition from $$n = {n_1}$$  to $$n = {n_2}.$$  The time period of electron in the initial state is eight times that in the final state. Then which of the following statement is true ?

Solution :
In the n^th orbit, let $${r_n}$$ be the radius and $${v_n}$$ be the speed of electron.
Time period, $${T_n} = \frac{{2\pi {r_n}}}{{{v_n}}} \propto \frac{{{r_n}}}{{{v_n}}}$$
Now $${r_n} \propto {n^2};{v_n} \propto \frac{1}{n}$$
$$\therefore \frac{{{r_n}}}{{{v_n}}} \propto {n^3}\,\,{\text{or}}\,\,{T_n} \propto {n^3}$$
Here $$8 = {\left( {\frac{{{n_1}}}{{{n_2}}}} \right)^3}\,\,{\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = 2$$
i.e., $${n_1} = 2{n_2}$$