Question

Angular momentum of the electron in a hydrogen atom is $\frac{3h}{2\pi}$, then find the energy of the electron in the orbit :-

• A. $-13.6 \text{ eV}$
• B. $-3.4 \text{ eV}$
• C. $-1.51 \text{ eV}$
• D. $-0.85 \text{ eV}$

Answer 1

The Correct Option is C

The angular momentum of an electron in a hydrogen atom is quantized and is given by Bohr's quantization condition:

\(L = \frac{nh}{2\pi}\), where \(n\) is the principal quantum number.

From the given problem, it is stated that the angular momentum of the electron is:

\(\frac{3h}{2\pi}\)

Comparing this with Bohr's quantization condition:

\(\frac{3h}{2\pi} = \frac{nh}{2\pi}\)

We can cancel \(\frac{h}{2\pi}\) from both sides, leading to:

\(n = 3\)

The energy of the electron in the nth orbit of the hydrogen atom is given by:

\(E_n = -\frac{13.6}{n^2} \text{ eV}\)

Substituting \(n = 3\) into the formula:

\(E_3 = -\frac{13.6}{3^2} \text{ eV} = -\frac{13.6}{9} \text{ eV} = -1.51 \text{ eV}\)

Thus, the correct energy of the electron in the orbit with \(n = 3\) is \(-1.51 \text{ eV}\).

The correct answer is:

$-1.51 \text{ eV}$