Question

The difference between the radii of $n^{th}$ and $(n +1)^{th}$ orbits of hydrogen atom is equal to the radius of $(n -1)^{th}$ orbit of hydrogen. The angular momentum of the electron in the $n^{th}$ orbit is ________ ($h$ is Plank�s constant)

• A. $\frac{h}{\pi}$
• B. $\frac{2h}{\pi}$
• C. $\frac{3h}{\pi}$
• D. $\frac{4h}{\pi}$

Answer 1

The Correct Option is B

Radius of $n^{\text {th}}$ orbit in an atom,
$r_{n}=\frac{n^{2} h^{2}}{4 \pi^{2} m Z e^{2}}$
So, $r_{n} \propto n^{2}$
Given in question, difference between radii of $(n+1)^{\text {th }}$ and $n^{\text {th }}$ orbit = radius of $n^{\text {th }}$ orbit.
$\Rightarrow (n+1)^{2}-n^{2} =(n-1)^{2}$
$n^{2}-4 n =0$ or $n=4$
According to Bohr's postulate, angular momentum is an integral multiple of $\frac{h}{2 \pi}$.
So, for $4^{\text {th }}$ orbit,
$L=\frac{n h}{2 \pi}=\frac{4 h}{2 \pi}=\frac{2 h}{\pi}$