Question

The energy of an electron in a hydrogen atom in ground state is -13.6 eV. Its energy in an orbit corresponding to quantum number \( n \) is -0.544 eV. The value of \( n \) is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

The energy of an electron in a hydrogen atom is inversely proportional to the square of the quantum number \( n \).

Answer 1

The Correct Option is D

The energy of an electron in a hydrogen atom is described by \(E_n = -\frac{13.6}{n^2} \, \text{eV}\), where \(E_n\) is the energy and \(n\) is the principal quantum number.

Given an electron energy of -0.544 eV, we solve for \(n\) using the equation: \(-\frac{13.6}{n^2} = -0.544\).

Simplifying, we get: \(\frac{13.6}{n^2} = 0.544\).

Rearranging to solve for \(n^2\): \(n^2 = \frac{13.6}{0.544}\).

Calculating the value: \(n^2 = 25\).

Taking the square root yields: \(n = 5\).

Therefore, the quantum number corresponding to an energy of -0.544 eV is 5.