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:
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:
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To solve these quickly: if the orbit number doubles (from 2 to 4), the energy becomes $\frac{1}{2^2} = \frac{1}{4}$ of its previous value. Conversely, if the orbit number triples, the energy becomes $\frac{1}{9}$ of the original value. Always remember that energy levels get closer together as $n$ increases.
Updated On: May 13, 2026
- $-41 \, \text{kJ/mol}$
- $-82 \, \text{kJ/mol}$
- $-164 \, \text{kJ/mol}$
- $-1312 \, \text{kJ/mol}$
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The Correct Option is B
Solution and Explanation
Concept:
According to Bohr's theory of the hydrogen atom, the energy of an electron in the $n^{th}$ orbit ($E_n$) is inversely proportional to the square of the principal quantum number ($n$). The mathematical expression for the energy of an orbit is:
$$E_n = -\frac{k \cdot Z^2}{n^2}$$
Where:
• $E_n$ is the energy of the $n^{th}$ orbit.
• $Z$ is the atomic number (for Hydrogen, $Z = 1$).
• $n$ is the principal quantum number (orbit number).
• $k$ is a constant representing the energy of the ground state. From this formula, we can derive a proportionality relationship for a specific element: $$E_n \propto \frac{1}{n^2}$$
Step 1: Establishing the ratio between two orbits. Let $E_2$ be the energy of the second orbit ($n_1 = 2$) and $E_4$ be the energy of the fourth orbit ($n_2 = 4$). Based on the inverse-square relationship: $$\frac{E_4}{E_2} = \frac{n_1^2}{n_2^2}$$
Substituting the orbit numbers:
$$\frac{E_4}{E_2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}$$
Step 2: Substituting the given value. We are given that the energy of the second Bohr orbit ($E_2$) is $-328 \, \text{kJ mol}^{-1}$. Using the relationship derived in
Step 1: $$E_4 = E_2 \times \frac{1}{4}$$ $$E_4 = -328 \times \frac{1}{4}$$
Step 3: Final Calculation. Perform the division: $$E_4 = - \frac{328}{4}$$ $$E_4 = -82 \, \text{kJ mol}^{-1}$$ This calculation shows that as the electron moves further from the nucleus (to a higher orbit), the energy level increases (becomes less negative).
• $E_n$ is the energy of the $n^{th}$ orbit.
• $Z$ is the atomic number (for Hydrogen, $Z = 1$).
• $n$ is the principal quantum number (orbit number).
• $k$ is a constant representing the energy of the ground state. From this formula, we can derive a proportionality relationship for a specific element: $$E_n \propto \frac{1}{n^2}$$
Step 1: Establishing the ratio between two orbits. Let $E_2$ be the energy of the second orbit ($n_1 = 2$) and $E_4$ be the energy of the fourth orbit ($n_2 = 4$). Based on the inverse-square relationship: $$\frac{E_4}{E_2} = \frac{n_1^2}{n_2^2}$$
Substituting the orbit numbers:
$$\frac{E_4}{E_2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}$$
Step 2: Substituting the given value. We are given that the energy of the second Bohr orbit ($E_2$) is $-328 \, \text{kJ mol}^{-1}$. Using the relationship derived in
Step 1: $$E_4 = E_2 \times \frac{1}{4}$$ $$E_4 = -328 \times \frac{1}{4}$$
Step 3: Final Calculation. Perform the division: $$E_4 = - \frac{328}{4}$$ $$E_4 = -82 \, \text{kJ mol}^{-1}$$ This calculation shows that as the electron moves further from the nucleus (to a higher orbit), the energy level increases (becomes less negative).
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