Question

The total energy of an electron in the $n^{th}$ stationary orbit of the hydrogen atom can be obtained by

• A. $E_{n}=\frac{13.6}{n^{2}} eV$
• B. $E_{n}=-\frac{13.6}{n^{2}} eV$
• C. $E_{n}=-\frac{1.36}{n^{2}}eV$
• D. $E_{n}=-13.6\times n^{2}eV$

Answer 1

The Correct Option is B

To solve the given question, we need to understand the concept of the total energy of an electron in the \(n^\text{th}\) stationary orbit of a hydrogen atom. According to quantum mechanics, the energy levels of an electron in a hydrogen atom are quantized.

The energy of an electron in the \(n^\text{th}\) orbit is given by the formula:

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

where:

• \(E_n\) is the energy of the electron in the \(n^\text{th}\) orbit.
• \(13.6 \text{ eV}\) is the energy of the electron in the ground state (\(n=1\)).
• \(n\) is the principal quantum number, representing the orbit number.

This formula shows that the energy is inversely proportional to the square of the principal quantum number \(n\). The negative sign indicates that the energy is bound, meaning the electron is attracted to the nucleus and possesses lower energy compared to a free electron.

Let's evaluate the given options based on the formula above:

• \(E_{n}=\frac{13.6}{n^{2}} \text{ eV}\) - Incorrect because it lacks a negative sign indicating bound state energy.
• \(E_{n}=-\frac{13.6}{n^{2}} \text{ eV}\) - Correct because it matches the derived formula.
• \(E_{n}=-\frac{1.36}{n^{2}} \text{ eV}\) - Incorrect, as it uses the wrong constant value.
• \(E_{n}=-13.6 \times n^{2} \text{ eV}\) - Incorrect, as it incorrectly assumes a direct relationship with \(n^{2}\).

Therefore, the correct option is \(E_{n}=-\frac{13.6}{n^{2}} \text{ eV}\), which accurately represents the energy of an electron in the \(n^\text{th}\) orbit of a hydrogen atom.