Question

The angular momentum of an electron in hydrogen atom is $\frac{4h}{2\pi }$. Kinetic energy of this electron is

• A. $4.35eV$
• B. $1.51eV$
• C. $0.85eV$
• D. $13.6eV$

Answer 1

The correct option is C $0.85eV$

Given, angular momentum is $\frac{4h}{2\pi }$
Therefore, $n=4$
$r_4=0.529\times 16=8.464\mathring{A}$
$K.E=\frac{L^2}{2m{r}^2}$

Mass of electron is $9.1\times {10}^{-31}kg$
Therefore, Kinetic energy is $=\frac{{\left(\frac{4\times 6.625\times {10}^{-34}}{2\times 3.14}\right)}^2}{2(9.1\times {10}^{-31}){(8.464\times {10}^{-10})}^2}=1.365\times {10}^{-15}J$
i.e., Kinetic energy is $=0.85eV$