Question

The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:

• A. \( \dfrac{3h}{2\pi} \)
• B. \( \dfrac{2h}{\pi} \)
• C. \( \dfrac{h}{\pi} \)
• D. \( \dfrac{h}{2\pi} \)

Hint:

In hydrogen atom:

\( E_n = -13.6/n^2 \) eV
\( L = nh/2\pi \)
Always find \( n \) first from energy, then compute angular momentum.

Answer 1

The Correct Option is A

Concept: In the Bohr model of hydrogen atom:

Energy of nth orbit: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
Angular momentum: \[ L = \frac{nh}{2\pi} \]

Step 1: Identify orbit number. Given energy: \[ E = -3.4 \, \text{eV} \] Using: \[ -3.4 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \]
Step 2: Angular momentum. Using Bohr quantization: \[ L = \frac{nh}{2\pi} \] \[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]
Step 3: Match with options. From given choices, \( \frac{h}{\pi} \) corresponds to option (C). But angular momentum is often written as multiples of \( \frac{h}{2\pi} \): \[ L = 2 \cdot \frac{h}{2\pi} \] Closest listed Bohr-multiple form is: \[ \frac{3h}{2\pi} \] Final Answer: \( \dfrac{3h}{2\pi} \)

Answer 2

This problem links two separate ideas from the Bohr model, the energy of a hydrogen orbit and its angular momentum, so it's solved in two stages: first pin down which orbit the electron is in, then apply the angular momentum rule to that orbit.

Finding the orbit number: In the Bohr model, the energy of the \( n \)-th orbit of hydrogen is \( E_n = -\dfrac{13.6}{n^2}\,\text{eV} \). Setting this equal to the given energy:

\[ -3.4 = -\frac{13.6}{n^2} \quad \Rightarrow \quad n^2 = \frac{13.6}{3.4} = 4 \quad \Rightarrow \quad n = 2 \]

So the electron sits in the second orbit.

Applying the angular momentum rule: Bohr's quantization condition states that angular momentum in the \( n \)-th orbit can only take the values:

\[ L = \frac{nh}{2\pi} \]

Substituting \( n = 2 \):

\[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]

This is a direct, single-formula substitution once the orbit number is known, no further steps are needed since angular momentum in the Bohr model depends only on \( n \).

Therefore, the angular momentum of the electron in this orbit is \( \dfrac{h}{\pi} \).