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 $1.51\text{eV}$

According to Bohr's postulate, the angular momentum of the electron in the stationary states is given by,

$L=\frac{nh}{2\pi }.........\left(1\right)$

Given that, $L=\frac{3h}{2\pi }.........\left(2\right)$

On comparing $\left(1\right)$ & $\left(2\right)$, we get,

$\Rightarrow n=3$

Total energy of the stationary states is given by,

$E=-\frac{13.6}{n^2}\text{eV}$

$\Rightarrow E=-\frac{13.6}{3^2}\text{eV}=-1.51\text{eV}$

From the relation,

$K.E=-E=-\frac{P.E}{2}$

$K.E=1.51\text{eV}$

Hence, option $\left(\text{B}\right)$ is correct.