Question

The magnitude of energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number $n$ are $E,P$ and $r$ respectively. Then, according to Bohr's theory, $EPr$ is proportional to

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

Answer 1

The correct option is A $\frac{1}{n}$

Energy of electron in $n^{\text{th}}$ orbit is
$E=-\frac{2{\pi }^2{K}^2{Z}^2{e}^4{m}_e}{n^2{h}^2}$
Momentum of electron in $n^{\text{th}}$ orbit is
$P=m{V}_n=\frac{nh}{2\pi r}$
Where $r$ is given as
$r=\frac{n^2{h}^2}{2\pi KZ{e}^2{m}_2}$
So,
$E\propto \frac{1}{n^2}$
$P\propto \frac{1}{n}$
$r\propto n^2$

hence, $EPr\propto \frac{1}{n}$