Question

An electron in a hydrogen-like atom has energy equal to \( -0.04 E_0 \), where \( E_0 \) is the magnitude of energy of this electron in ground state in eV. If the angular momentum of this electron is \( L \), then the value of \( \frac{2 \pi L}{h} \) is (where \( h \) is Planck's constant).

• A. 1
• B. 4
• C. 5
• D. 6

Hint:

The angular momentum of an electron in a hydrogen-like atom is quantized and is given by \( L = n \hbar \), where \( n \) is the principal quantum number.

Answer 1

The Correct Option is C

Step 1: Understanding the energy levels.
The energy of the electron in a hydrogen-like atom is given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] where \( n \) is the principal quantum number. The energy of the electron in the ground state is \( E_1 = -13.6 \, \text{eV} \).
Step 2: Given energy relation.
The problem states that the energy of the electron is \( -0.04 E_0 \), where \( E_0 \) is the magnitude of the ground state energy. Therefore: \[ E = -0.04 E_0 = -0.04 \times 13.6 = -0.544 \, \text{eV} \] Now, using the formula for the energy levels of hydrogen-like atoms: \[ E_n = - \frac{13.6}{n^2} \] Equating the two expressions for energy, we get: \[ - \frac{13.6}{n^2} = -0.544 \] Solving for \( n \): \[ \frac{13.6}{n^2} = 0.544 \] \[ n^2 = \frac{13.6}{0.544} = 25 \] \[ n = 5 \]
Step 3: Angular momentum.
For an electron in the \( n \)-th orbit, the angular momentum is given by: \[ L = n \hbar \] where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant. So, the angular momentum \( L \) for \( n = 5 \) is: \[ L = 5 \times \hbar = 5 \times \frac{h}{2\pi} \]
Step 4: Calculate \( \frac{2 \pi L}{h} \).
Substitute \( L = 5 \times \frac{h}{2\pi} \) into \( \frac{2 \pi L}{h} \): \[ \frac{2 \pi L}{h} = \frac{2 \pi \times 5 \times \frac{h}{2 \pi}}{h} = 5 \] Thus, the value of \( \frac{2 \pi L}{h} \) is 5.