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 given by the formula:

\(E_n = -\frac{13.6}{n^2} \, \text{eV}\)

where \(E_n\) is the energy of the electron and \(n\) is the principal quantum number.

Given that the energy of the electron is -0.544 eV, we can set up the equation:

\(-\frac{13.6}{n^2} = -0.544\)

Removing the negative signs, we have:

\(\frac{13.6}{n^2} = 0.544\)

Rearrange the equation to solve for \(n^2\):

\(n^2 = \frac{13.6}{0.544}\)

Calculate the value:

\(n^2 = 25\)

Taking the square root on both sides, we find:

\(n = 5\)

Thus, the quantum number \(n\) corresponding to the energy of -0.544 eV is 5.