Question:

The magnitude of total energy and angular momentum of an electron in the \( n \)th orbit of a Bohr atom is denoted by \( E_n \) and \( L_n \), respectively. Then

Show Hint

In Bohr's model, the energy and angular momentum of an electron in orbit are related to the quantum number \( n \).
Updated On: Jan 26, 2026
  • \( E_n \propto L_n \)
  • \( E_n \propto L_n^3 \)
  • \( E_n \propto \frac{1}{L_n^2} \)
  • \( E_n \propto \frac{1}{L_n} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Step 1: Understanding the energy and angular momentum relation.
In Bohr's model of the atom, the angular momentum of an electron is quantized and given by \( L_n = n h / 2 \pi \), where \( n \) is the principal quantum number. The total energy of the electron is given by \( E_n = - \frac{13.6}{n^2} \, \text{eV} \).
Step 2: Deriving the relationship.
We know that \( L_n \propto n \) and \( E_n \propto \frac{1}{n^2} \). Thus, we can derive the relation between \( E_n \) and \( L_n \) as: \[ E_n \propto \frac{1}{L_n^2} \] Step 3: Conclusion.
The correct answer is (C), \( E_n \propto \frac{1}{L_n^2} \).
Was this answer helpful?
0
0

Top Questions on Atomic Physics

View More Questions