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 $1.51\text{eV}$

Given, $L=\frac{3h}{2\pi }$

According to Bohr’s $2^{nd}$ Postulate-

$L=\frac{nh}{2\pi }\Rightarrow n=3$

$\because$ Total energy of $H-$ atom is,

$TE=-\frac{13.6Z^2}{n^2}$

$TE=\frac{-13.6(1{)}^2}{3^2}$

$\because KE=-TE=13.6\times \frac{1}{9}=1.51\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.