The energy, radius and velocity of the electron in the hydrogen atom in the ground state are `-13.6 eV, 0.53 Å` and `2.188 xx 10^(8) m s^(-1)` respectively.
The energy of the electron in the third orbit will be

Options

A. `-4.4 eV`
B. `-1.5 eV`
C. `-40.8eV`
D. `-22.4 eV`

Explanation

To find the energy of the electron in the third orbit of a hydrogen atom, we can use the formula for the energy levels of an electron in a hydrogen atom, which is given by:

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

where \(E_n\) is the energy of the electron in the nth orbit and \(n\) is the principal quantum number.

### Step-by-Step Solution:

1. **Identify the Principal Quantum Number**:
- For the third orbit, the principal quantum number \(n\) is 3.

2. **Substitute the Value of \(n\) into the Energy Formula**:
- We will substitute \(n = 3\) into the energy formula:
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} \]

3. **Calculate \(3^2\)**:
- First, calculate \(3^2\):
\[ 3^2 = 9 \]

4. **Substitute \(3^2\) into the Energy Equation**:
- Now substitute \(9\) into the equation:
\[ E_3 = -\frac{13.6 \, \text{eV}}{9} \]

5. **Perform the Division**:
- Now, calculate the value:
\[ E_3 = -1.5111 \, \text{eV} \approx -1.5 \, \text{eV} \]

### Final Answer:
The energy of the electron in the third orbit is approximately \(-1.5 \, \text{eV}\).

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