Question:
If the energy of a hydrogen atom in \(n^{\text{th}}\) orbit is \(E_n\) then energy in the \(n^{\text{th}}\) orbit of a singly ionized helium atom will be
If the energy of a hydrogen atom in \(n^{\text{th}}\) orbit is \(E_n\) then energy in the \(n^{\text{th}}\) orbit of a singly ionized helium atom will be
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Energy is proportional to \(Z^2\). For He\(^+\) (\(Z=2\)), energy is 4 times that of H (\(Z=1\)) for same n.
Updated On: Apr 23, 2026
- \(4E_n\)
- \(E_n/4\)
- \(2E_n\)
- \(E_n/2\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Energy of hydrogen-like atom: \(E = -\frac{13.6 Z^2}{n^2}\) eV. For H, \(Z=1\); for He\(^+\), \(Z=2\).
Step 2: Detailed Explanation:
\(E_n(\text{H}) = -\frac{13.6 \times 1^2}{n^2}\).
\(E_n(\text{He}^+) = -\frac{13.6 \times 2^2}{n^2} = -\frac{13.6 \times 4}{n^2} = 4 \times E_n(\text{H})\).
Step 3: Final Answer:
Thus, energy = \(4E_n\).
Energy of hydrogen-like atom: \(E = -\frac{13.6 Z^2}{n^2}\) eV. For H, \(Z=1\); for He\(^+\), \(Z=2\).
Step 2: Detailed Explanation:
\(E_n(\text{H}) = -\frac{13.6 \times 1^2}{n^2}\).
\(E_n(\text{He}^+) = -\frac{13.6 \times 2^2}{n^2} = -\frac{13.6 \times 4}{n^2} = 4 \times E_n(\text{H})\).
Step 3: Final Answer:
Thus, energy = \(4E_n\).
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