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

Answer: (B)

Radius of $n^{\text{th}}$ orbit in an atom,
$r_n=\frac{n^2{h}^2}{4{\pi }^{2}mZ{e}^2}$
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.
$\Rightarrow (n+1{)}^2-n^2=(n-1{)}^2$
$n^2-4n=0$ 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,
$L=\frac{nh}{2\pi }=\frac{4h}{2\pi }=\frac{2h}{\pi }$