An electron of a hydrogen like atom, having Z= 4, jumps from `4^th` energy state to `2^nd` energy state. The energy released in this process, will be
(Given Rch = 13.6 eV)
Where R = Rydberg constant, c = speed of light in vacuum, h = Planck's constant.

Options

A. `40.8 eV`
B. `3.4 eV`
C. `10.5 eV`
D. `13.6 eV`

Answer

To solve the problem, we need to calculate the energy released when an electron in a hydrogen-like atom with atomic number \( Z = 4 \) transitions from the \( n = 4 \) energy state to the \( n = 2 \) energy state.

### Step-by-Step Solution:

1. **Identify the Formula for Energy Levels**:
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( E_n \) is the energy at the principal quantum number \( n \), and \( Z \) is the atomic number.

2. **Calculate the Energy at \( n = 4 \)**:
For \( n_1 = 4 \):
\[ E_1 = -\frac{13.6 \times 4^2}{4^2} = -\frac{13.6 \times 16}{16} = -13.6 \, \text{eV} \]

3. **Calculate the Energy at \( n = 2 \)**:
For \( n_2 = 2 \):
\[ E_2 = -\frac{13.6 \times 4^2}{2^2} = -\frac{13.6 \times 16}{4} = -54.4 \, \text{eV} \]

4. **Calculate the Energy Released (\( \Delta E \))**:
The energy released when the electron jumps from \( n_1 \) to \( n_2 \) is given by:
\[ \Delta E = E_2 - E_1 \]
Substituting the values we calculated:
\[ \Delta E = (-54.4) - (-13.6) = -54.4 + 13.6 = -40.8 \, \text{eV} \]
The negative sign indicates that energy is released, so we take the absolute value:
\[ \Delta E = 40.8 \, \text{eV} \]

### Final Answer:
The energy released in this process is \( 40.8 \, \text{eV} \).