In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.

Options

A. ` + 3.4 eV`
B. `+ 6.8 eV`
C. `-13.6 eV`
D. `+ 13.6 eV`

Explanation

To find the kinetic energy (KE) of the hydrogen atom in its first excited state (n=2), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels of Hydrogen Atom**:
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Determine the Energy at n=2**:
For the first excited state, \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
This confirms that the energy of the first excited state is indeed \(-3.4 \, \text{eV}\).

3. **Relate Total Energy to Kinetic Energy**:
In a hydrogen atom, the total energy (E) is related to the kinetic energy (KE) by the equation:
\[ E = KE + PE \]
where PE is the potential energy. For a hydrogen atom, the potential energy is twice the negative of the kinetic energy:
\[ PE = -2 \cdot KE \]

4. **Substituting the Total Energy**:
Since the total energy \( E \) is equal to \(-3.4 \, \text{eV}\):
\[ -3.4 \, \text{eV} = KE - 2 \cdot KE \]
Simplifying this gives:
\[ -3.4 \, \text{eV} = -KE \]

5. **Solving for Kinetic Energy**:
Rearranging the equation gives:
\[ KE = 3.4 \, \text{eV} \]

### Final Answer:
The kinetic energy of the hydrogen atom in the first excited state (n=2) is \( \mathbf{3.4 \, eV} \).