Question

The angular momentum of an electron in hydrogen atom is $\frac{h}{\pi }$ . The kinetic energy of the electron is

• A. 13.6 eV
• B. 3.4 eV
• C. 1.51 eV
• D. 0.85 eV

Answer 1

Answer: (B)

The angular momentum of an electron in hydrogen atom $=\frac{h}{\pi }$
As we know, from Bohr’s second postulate,
$mvr=\frac{h}{\pi }\Rightarrow v=\frac{h}{\pi mr}$...(i)
Now, kinetic energy of electron
$=\frac{1}{2}m{v}^2=\frac{1}{2}m\times {\left(\frac{h}{\pi mr}\right)}^2$
[from Eq. $\left(i\right)$]
$=\frac{h^2}{2{\pi }^{2}m{r}^2}$
$=\frac{{\left(6.63\times 10^{-34}\right)}^2}{2\times {\left(3.14\right)}^2\times 9.1\times 10^{-31}\times {\left(5.3\times 10^{-11}\right)}^2\times 1.6\times 10^{-19}}$
$=3.4eV$