The minimum energy that must be given to a H atom in ground state so that it can emit an H, line in balmer series is

Options

A. `12.4eV`
B. `10.2eV`
C. `13.06eV`
D. `13.6eV`

Explanation

To find the minimum energy that must be given to a hydrogen atom in the ground state so that it can emit a line in the Balmer series, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The lines in this series are visible light emissions.

2. **Identify the Transition for the Minimum Energy**: The minimum energy transition in the Balmer series corresponds to the transition from n = 5 to n = 2. This is because the higher the initial energy level, the more energy is required to excite the electron.

3. **Calculate the Energy Levels**: The energy levels 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.

4. **Calculate the Energy for n = 5**:
\[ E_5 = -\frac{13.6 \, \text{eV}}{5^2} = -\frac{13.6 \, \text{eV}}{25} = -0.544 \, \text{eV} \]

5. **Calculate the Energy for n = 2**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

6. **Calculate the Energy Difference**: The energy difference (ΔE) required to excite the electron from n = 1 to n = 5 is given by:
\[ \Delta E = E_5 - E_1 \]
where \( E_1 = -13.6 \, \text{eV} \) (ground state energy).
\[ \Delta E = (-0.544 \, \text{eV}) - (-13.6 \, \text{eV}) = 13.06 \, \text{eV} \]

7. **Conclusion**: The minimum energy that must be given to the hydrogen atom in the ground state to emit a line in the Balmer series (specifically the H gamma line) is **13.06 eV**.