Question

The energy of an electron in the $nth$ Bohr orbit of H - atom is

• A. $\frac{-13.6}{ n ^{2}} eV$
• B. $\frac{-13.6}{ n } eV$
• C. $\frac{-13.6}{ n ^{4}} eV$
• D. $\frac{-13.6}{n^{3}} e V$

Answer 1

The Correct Option is A

To determine the energy of an electron in the $nth$ Bohr orbit of a hydrogen atom, we use the formula for the energy levels in hydrogen-like atoms:

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

Here's the step-by-step reasoning for this formula:

1. According to Bohr's model, the energy levels of the hydrogen atom are quantized. This means that electrons can only exist at specific energy levels.

2. The energy of the $nth$ level is derived using the following expression:

   E_n = -\frac{me^4}{8\epsilon_0^2h^2n^2}

   Where:

   • m is the mass of the electron
   • e is the charge of the electron
   • \epsilon_0 is the permittivity of free space
   • h is Planck's constant
   • n is the principal quantum number (Bohr orbit number)

3. For hydrogen, this simplifies to:

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

   This indicates that the energy of an electron becomes less negative and closer to zero as the principal quantum number $n$ increases. This signifies that as the electron occupies higher orbits, it requires less energy to be completely freed from the atom.

Therefore, the correct option for the energy of an electron in the $nth$ Bohr orbit of the hydrogen atom is:

\frac{-13.6}{n^2} \text{ eV}

This matches with the option: \frac{-13.6}{n^2} \, \text{eV}