The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV`. The possible energy value (s) of the excited state (s) for electron in Bohr orbits of hydrogen is (are)

Options

A. `-3.4 eV`
B. `-4.2eV`
C. `-6.8 eV`
D. `-1.5 eV`

Explanation

To find the possible energy values of the excited states for an electron in the Bohr orbits of a hydrogen atom, we can use the formula derived from Bohr's model of the hydrogen atom. The energy of an electron in the nth orbit is given by the formula:

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

where:
- \( E_n \) is the energy of the electron in the nth orbit,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number (n = 1, 2, 3, ...).

### Step-by-Step Solution:

1. **Identify the Formula**:
We start with the energy formula for the hydrogen atom:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

2. **Calculate Energy for Different Excited States**:
We will calculate the energy for different values of \( n \) (where \( n \) can be 2, 3, 4, etc. for excited states).

- For \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

- For \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

- For \( n = 4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

3. **List Possible Energy Values**:
The possible energy values for the excited states (n = 2, 3, 4, ...) are:
- For \( n = 2 \): \( -3.4 \, \text{eV} \)
- For \( n = 3 \): \( -1.51 \, \text{eV} \)
- For \( n = 4 \): \( -0.85 \, \text{eV} \)

4. **Conclusion**:
The possible energy values of the excited states for the electron in the Bohr orbits of hydrogen are \( -3.4 \, \text{eV} \), \( -1.51 \, \text{eV} \), and \( -0.85 \, \text{eV} \).