A hydrogen atom in ground state is given an energy of 10.2 eV. How many spectral lines will be emitted due to transition of electrons ?
- 6
- 3
- 10
- 1
The Correct Option is D
Approach Solution - 1
\[ \therefore E \propto \frac{Z^2}{n^2} \] Energy of the 2^{nd} orbit: \[ E_2 = \frac{-13.6}{4} = -3.4\, \text{eV} \] \[ \Delta E_{2 \to 1} = 10.2\, \text{eV} \] So, only possible transition is from: \[ n = 2 \text{ to } n = 1 \]
Approach Solution -2
To determine the number of spectral lines emitted when an electron in a hydrogen atom transitions after being excited, we first need to understand the given conditions. A hydrogen atom in the ground state (\(n=1\)) is excited with an energy of 10.2 eV.
The energy difference between the ground state and the first excited state (\(n=2\)) in a hydrogen atom is approximately 10.2 eV. When an electron is given exactly 10.2 eV of energy, it transitions from \(n=1\) to \(n=2\).
Once the electron is in the excited state \(n=2\), it can only transition back to \(n=1\) because there are no other lower energy levels that correspond to the given energy input. The transition thus results in the emission of energy in the form of a photon, corresponding to a single spectral line.
Based on this understanding, the number of spectral lines emitted is:
- 1 spectral line from the transition \(n=2 \rightarrow n=1\).
Hence, the correct option is 1.
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