Question:
A hydrogen atom is excited to level \(n\) where its potential energy is found to be \(-1.088\, \text{eV}\). The magnitude of \(n\) is:
A hydrogen atom is excited to level \(n\) where its potential energy is found to be \(-1.088\, \text{eV}\). The magnitude of \(n\) is:
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In hydrogen atom: \(U = 2E\) and \(E \propto -1/n^2\), always useful shortcut for level finding.
Updated On: Jun 20, 2026
- 5
- 4
- 6
- 3
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The Correct Option is A
Solution and Explanation
Step 1: Use hydrogen atom energy relations.
Total energy of hydrogen atom: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] Potential energy is: \[ U = 2E_n \]
Step 2: Write expression for potential energy.
\[ U = -\frac{27.2}{n^2} \, \text{eV} \] Given: \[ U = -1.088 \, \text{eV} \]
Step 3: Equate values.
\[ \frac{27.2}{n^2} = 1.088 \]
Step 4: Solve for \(n^2\).
\[ n^2 = \frac{27.2}{1.088} \] \[ n^2 = 25 \]
Step 5: Find \(n\).
\[ n = 5 \]
Step 6: Final conclusion.
\[ \boxed{5} \]
Total energy of hydrogen atom: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] Potential energy is: \[ U = 2E_n \]
Step 2: Write expression for potential energy.
\[ U = -\frac{27.2}{n^2} \, \text{eV} \] Given: \[ U = -1.088 \, \text{eV} \]
Step 3: Equate values.
\[ \frac{27.2}{n^2} = 1.088 \]
Step 4: Solve for \(n^2\).
\[ n^2 = \frac{27.2}{1.088} \] \[ n^2 = 25 \]
Step 5: Find \(n\).
\[ n = 5 \]
Step 6: Final conclusion.
\[ \boxed{5} \]
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