Question

The period of revolution of an electron revolving in $n^{th}$ orbit of $H$-atom is proportional to

• A. $n^2$
• B. $\frac{1}{n}$
• C. $n^3$
• D. Independent of n

Answer 1

The Correct Option is C

As we know that time period of revolution of an electron is
$T =2 \pi r / v$,
And $r \propto n^{2} / Z^{2}$ and $v \propto Z / n$
$\therefore T \propto n ^{3} / Z ^{2}$ and for $H$ -atom
$Z =1, T \propto n ^{3}$

Answer 2

In Bohr's model of the atom, the period \(T\) of revolution of an electron in the \(n^{\text{th}}\) orbit is related to the radius of the orbit. The radius \(r_n\) is proportional to \(n^2\) and the velocity \(v_n\) is inversely proportional to \(n\).
Since the period \(T\) is the time taken to complete one revolution, it is proportional to \(n^3\) because it depends on both the radius and the speed of the electron.

Therefore, the period of revolution is proportional to \(n^3\).