Question

The energy of the second bohr orbit of the hydrogen atom is $-328 \text{ kJ mol}^{-1}$; hence the energy of the fourth bohr orbit would be:

• A. $ -41 \text{ kJ/mol} $
• B. $ -82 \text{ kJ/mol} $
• C. $ -164 \text{ kJ/mol} $
• D. $ -1312 \text{ kJ/mol} $

Hint:

For hydrogen atom problems based on Bohr’s model, always remember: \[ E_n \propto \frac{1}{n^2} \] If the orbit number becomes twice, the energy becomes one-fourth. Here: \[ 2 \rightarrow 4 \] So, \[ E_4 = \frac{E_2}{4} \] which gives: \[ -328/4 = -82 \]

Answer 1

The Correct Option is B

Concept: According to Bohr’s atomic theory, the energy of an electron present in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the relation: \[ E_n = \frac{-13.6}{n^2} \text{ eV} \] or in general proportional form, \[ E_n \propto \frac{-1}{n^2} \] This means that the energy of an electron becomes less negative as the orbit number increases. The negative sign indicates that the electron is bound to the nucleus. For two different orbits of the same atom, we can write: \[ \frac{E_1}{E_2} = \frac{n_2^2}{n_1^2} \] or equivalently, \[ E_n \propto \frac{1}{n^2} \] Since the question provides the energy of the second orbit and asks for the energy of the fourth orbit, we directly apply Bohr’s energy relation.

Step 1: Writing the given information.
From the question: \[ E_2 = -328 \text{ kJ mol}^{-1} \] We need to calculate: \[ E_4 = ? \] where, \[ n_1 = 2 \] and \[ n_2 = 4 \]

Step 2: Using the Bohr orbit energy relation.
Using the proportional relation: \[ \frac{E_4}{E_2} = \frac{n_1^2}{n_2^2} \] Substituting the values: \[ \frac{E_4}{-328} = \frac{2^2}{4^2} \] \[ \frac{E_4}{-328} = \frac{4}{16} \] \[ \frac{E_4}{-328} = \frac{1}{4} \]

Step 3: Calculating the energy of the fourth orbit.
Multiplying both sides by $-328$: \[ E_4 = -328 \times \frac{1}{4} \] \[ E_4 = -82 \text{ kJ mol}^{-1} \]

Step 4: Final conclusion.
Therefore, the energy associated with the fourth Bohr orbit of hydrogen atom is: \[ \boxed{-82 \text{ kJ mol}^{-1}} \] Hence, the correct answer is: \[ \boxed{(b)\ -82 \text{ kJ/mol}} \]