Q84 / 120 · VITEEE 2015 · Physics

The angular momentum of an electron in Bohr's hydrogen atom whose energy is $-3.4 \mathrm{eV}$, is

A
$\frac{5 h}{2 \pi}$
B
$\frac{h}{2 \pi}$
Correct answer
C
$\frac{h}{\pi}$
D
$\frac{2 h}{3 \pi}$

Solution

Energy of electron in nth orbit of hydrogen atom $\begin{aligned} & E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV} \\ \Rightarrow & 3.4=-\frac{13.6}{n^{2}} \Rightarrow n^{2}=4 \end{aligned}$ or, $\mathrm{n}=2$ Angular momentum of electron $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi}$

Hence, the correct answer is option: 'C'.

Relevant Questions

What is the ground state energy of positronium? (The ground state energy of hydrogen is $-13.6 {eV}$)2022Bohr's Atomic Model
The ratio of the energies of the hydrogen atom in its first to second excited states is2019Bohr's Atomic Model
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited sta2018Bohr's Atomic Model
The ratio of radii of the first three Bohr orbits is2018Bohr's Atomic Model
A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be2017Bohr's Atomic Model
An electron makes a transition from an excited state to the ground state of a hydrogen - like atom. Then2017Bohr's Atomic Model