Question

The angular momentum of an electron in the hydrogen atom is $\frac{3h}{2\pi }$, here $h$ is Planck’s constant. The kinetic energy of this electron is

• A. $4.53\text{eV}$
• B. $1.51\text{eV}$
• C. $3.4\text{eV}$
• D. $6.8\text{eV}$

Answer 1

The correct option is B $1.51\text{eV}$

Angular momentum of electron in $n^{th}$ orbit is $L_n=\frac{nh}{2\pi }$
Given that $L=\frac{3h}{2\pi }$
So, $n=3$

Using total energy in $n^{th}$ orbit formula $E=\frac{-13.6Z^2}{n^2}$
$E=\frac{-13.6}{9}$
$E=-1.51\text{eV}$
We know that,
$\text{K.E.}=-\text{T.E.}$
$\text{K.E.}=-(-1.51)=1.51\text{eV}$