Minimum excitation potential of Bohr's first orbit hydrogen atom is

Options

A. 13.6 V
B. 3.4 V
C. 10.2 V
D. 3.6 V

Answer

To find the minimum excitation potential of Bohr's first orbit in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels in Bohr's Model**:
In Bohr's model of the hydrogen atom, the energy levels (or orbits) are quantized. The energy of the nth orbit is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Calculate the Energy of the First Orbit (n=1)**:
For the first orbit (n=1):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

3. **Calculate the Energy of the Second Orbit (n=2)**:
For the second orbit (n=2):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

4. **Determine the Energy Difference for Excitation**:
The minimum excitation potential is the energy required to move an electron from the first orbit (n=1) to the second orbit (n=2). This is calculated as:
\[ \Delta E = E_2 - E_1 \]
Substituting the values we calculated:
\[ \Delta E = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = -3.4 \, \text{eV} + 13.6 \, \text{eV} = 10.2 \, \text{eV} \]

5. **Conclusion**:
The minimum excitation potential required to excite the electron from the first orbit to the second orbit in a hydrogen atom is:
\[ \text{Minimum Excitation Potential} = 10.2 \, \text{eV} \]

### Final Answer:
The minimum excitation potential of Bohr's first orbit hydrogen atom is **10.2 eV**.

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