Question:medium

Ionisation potential of hydrogen atom is \(13.6 \, \text{eV}\). Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy \(12.1 \, \text{eV}\). The spectral lines emitted by hydrogen atoms according to Bohr's theory will be:

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If electron reaches level \(n\), total spectral lines = \(\frac{n(n-1)}{2}\).
Updated On: Apr 17, 2026
  • one
  • two
  • three
  • four
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
When an atom absorbs energy, its electron transitions to a higher energy level. During de-excitation, it can transition through multiple steps back to the ground state, emitting photons of different frequencies (spectral lines).
Step 2: Key Formula or Approach:
1. Energy of nth level: \(E_{n} = \frac{-13.6}{n^{2}}\text{ eV}\).
2. Number of spectral lines: \(N = \frac{n(n - 1)}{2}\).
Step 3: Detailed Explanation:
The ground state energy of hydrogen is \(-13.6\text{ eV}\).
Excitation energy given \(= 12.1\text{ eV}\).
The energy of the excited state is:
\[ E_{final} = E_{1} + 12.1 = -13.6 + 12.1 = -1.5\text{ eV} \]
Check which orbit \(n\) corresponds to this energy:
\[ \frac{-13.6}{n^{2}} = -1.5 \implies n^{2} = \frac{13.6}{1.5} \approx 9 \implies n = 3 \]
The electrons are excited to the \(n = 3\) orbit.
Total spectral lines emitted for \(n = 3\):
\[ N = \frac{3(3 - 1)}{2} = \frac{3 \times 2}{2} = 3 \]
The three lines correspond to transitions: \(3 \to 1\), \(3 \to 2\), and \(2 \to 1\).
Step 4: Final Answer:
The number of spectral lines emitted is three.
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