A hydrogen atom in ground state is given an energy of $10.2 \mathrm{~eV}$. How many spectral lines will be emitted due to transition of electrons?

<p>To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2 \, \text{eV}$, we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to the ground state.</p> <p><b>Energy Levels of Hydrogen Atom :</b></p> <p>The energy levels $ E_n $ of a hydrogen atom can be calculated using the formula:</p> <p>$ E_n = -\frac{13.6 \, \text{eV}}{n^2} $</p> <p>where $ n $ is the principal quantum number.</p> <p><b>Ground State Energy :</b></p> <p>The ground state (n=1) energy is $ E_1 = -13.6 \, \text{eV} $.</p> <p><b>Determine the Excited State :</b></p> <p>If the ground state electron is given $10.2 \, \text{eV}$, its total energy becomes:</p> <p>$ E_{\text{total}} = E_1 + 10.2 \, \text{eV} = -13.6 \, \text{eV} + 10.2 \, \text{eV} = -3.4 \, \text{eV} $</p> <p>Now, we find the principal quantum number $ n $ for which the energy is closest to $-3.4 \, \text{eV}$:</p> <ul> <li>For $ n = 2 $: $ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} $</li><br> <li>For $ n = 3 $: $ E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} $</li> </ul> <p>Since $-3.4 \, \text{eV}$ matches exactly with $ E_2 $, the electron reaches the second energy level ($ n = 2 $).</p> <p><b>Counting Spectral Lines :</b></p> <p>When the electron falls back to the ground state from $ n = 2 $, it can do so in a single transition:</p> <ul> <li>$ n = 2 $ to $ n = 1 $</li> </ul> <p>Thus, only one spectral line will be emitted during this transition.</p> <p><b>Conclusion :</b></p> <p>The number of spectral lines emitted when a hydrogen atom in the ground state is given $10.2 \, \text{eV}$ and the electron transitions back to the ground state from $ n = 2 $ is just one.</p> <p>Therefore, the correct answer is:</p> <strong>Option D: 1</strong>.