QJEEM24Ph8cc4: JEE Main 2024 Physics - Class 12 JEEM24Ph8cc4 Solution: A hydrogen atom in ground state is given an energy of 10.2…

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.

Energy Levels of Hydrogen Atom :

The energy levels $E_n$ of a hydrogen atom can be calculated using the formula:

$E_n = -\frac{13.6 \, \text{eV}}{n^2}$

where $n$ is the principal quantum number.

Ground State Energy :

The ground state (n=1) energy is $E_1 = -13.6 \, \text{eV}$ .

Determine the Excited State :

If the ground state electron is given $10.2 \, \text{eV}$ , its total energy becomes:

$E_{\text{total}} = E_1 + 10.2 \, \text{eV} = -13.6 \, \text{eV} + 10.2 \, \text{eV} = -3.4 \, \text{eV}$

Now, we find the principal quantum number $n$ for which the energy is closest to $-3.4 \, \text{eV}$ :

• For $n = 2$ : $E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV}$


• For $n = 3$ : $E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV}$

Since $-3.4 \, \text{eV}$ matches exactly with $E_2$ , the electron reaches the second energy level ( $n = 2$ ).

Counting Spectral Lines :

When the electron falls back to the ground state from $n = 2$ , it can do so in a single transition:

• $n = 2$ to $n = 1$

Thus, only one spectral line will be emitted during this transition.

Conclusion :

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.

Therefore, the correct answer is: