Question

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 ?

• A. 6
• B. 3
• C. 10
• D. 1

Answer 1

The Correct Option is D

\[ \therefore E \propto \frac{Z^2}{n^2} \] 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: \[ n = 2 \text{ to } n = 1 \] 

Answer 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.