The difference between the radii of $n^{\text {th }}$ and $(n+1)^{\text {th }}$ orbits of hydrogen atom is…

The difference between the radii of $n^{\text {th }}$ and $(n+1)^{\text {th }}$ orbits of hydrogen atom is equal to the radius of $(n-1)^{\text {th }}$ orbit of hydrogen. The angular momentum of the electron in the $n^{\text {th }}$ orbit is _________ ( $h$ is Planck's constant).
  1. $\frac{h}{\pi}$
  2. $\frac{2h}{\pi}$
  3. $\frac{3h}{\pi}$
  4. $\frac{4h}{\pi}$

Solution

(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.
$\begin{aligned}
\Rightarrow \quad(n+1)^2-n^2 & =(n-1)^2 \\
n^2-4 n & =0 \text { or } n=4
\end{aligned}$
According to Bohr's postulate, angular momentum is an integral multiple of $\frac{h}{2 \pi}$.
So, for $4^{\text {th }}$ orbit,
$L=\frac{n h}{2 \pi}=\frac{4 h}{2 \pi}=\frac{2 h}{\pi}$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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