Question

Angular momentum of the electron in a hydrogen atom is $\frac{3h}{2\pi}$ then find total energy of electron (in eV/atom)

• A. –1.51
• B. –122.4
• C. –40.8
• D. –4.53

Answer 1

The Correct Option is A

Based on Bohr's model of the atom, the angular momentum of an electron in a stationary orbit is quantized, and the energy of the electron depends on the principal quantum number ($n$).
1. Determine the orbit number (n):
According to Bohr's postulate, Angular Momentum = $\frac{nh}{2\pi}$.
Given: $\frac{nh}{2\pi} = \frac{3h}{2\pi}$
Comparing both, we get $n = 3$.

2. Calculate Total Energy:
The formula for total energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is:
$E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV/atom}$
For Hydrogen, $Z = 1$.
For $n = 3$:
$E_3 = -13.6 \times \frac{1^2}{3^2} = -13.6 \times \frac{1}{9}$
$E_3 = -1.511... \text{ eV/atom}$

Rounding to two decimal places, the energy is –1.51 eV/atom. Option (1) is correct.