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:

For hydrogen atom: \[ E_n=-\frac{13.6}{n^2}\text{ eV} \] and \[ L=n\frac{h}{2\pi}. \] Always determine \(n\) from the energy first and then substitute into the angular momentum formula.

Answer 1

The Correct Option is A

Concept: According to Bohr's model of hydrogen atom, the energy of the electron in the \(n^{\text{th}}\) orbit is \[ E_n=-\frac{13.6}{n^2}\ \text{eV}. \] The angular momentum of the electron is quantized and is given by \[ L=n\frac{h}{2\pi}. \] Therefore, we first determine the principal quantum number \(n\) from the given energy and then calculate the angular momentum.

Step 1: Determine the orbit number corresponding to the given energy. Given, \[ E_n=-3.4\ \text{eV}. \] Using \[ -\frac{13.6}{n^2}=-3.4. \] Removing the negative sign, \[ \frac{13.6}{n^2}=3.4. \] Thus, \[ n^2=\frac{13.6}{3.4}=4. \] Hence, \[ n=2. \]

Step 2: Use Bohr's quantization condition. The angular momentum is \[ L=n\frac{h}{2\pi}. \] Substituting \(n=2\), \[ L=2\left(\frac{h}{2\pi}\right). \] Therefore, \[ L=\frac{h}{\pi}. \]

Step 3: Identify the correct option. Thus, \[ \boxed{L=\frac{h}{\pi}}. \] Hence, the correct answer is \[ \boxed{\text{(C)}}. \] Note: The option marked in some answer keys as \((A)\) is incorrect. Using the standard Bohr energy relation, \(-3.4\ \text{eV}\) corresponds to \(n=2\), giving \[ L=\frac{h}{\pi}. \]