If the potential energy of a H-atom in the ground state be zero then its potential energy in the first excited state will be

Options

A. 10.2ev
B. 20.4eV
C. 23.8eV
D. 27.2eV

Explanation

To find the potential energy of a hydrogen atom in the first excited state, we can use the formula for the potential energy of an electron in a hydrogen atom, which is given by:

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

where \( n \) is the principal quantum number corresponding to the energy level of the atom.

### Step 1: Identify the ground state and first excited state
- The ground state corresponds to \( n = 1 \).
- The first excited state corresponds to \( n = 2 \).

### Step 2: Calculate the potential energy in the ground state
For the ground state (\( n = 1 \)):
\[ PE_{ground} = -\frac{2 \times 13.6 \, \text{eV}}{1^2} = -27.2 \, \text{eV} \]
According to the question, this potential energy is considered to be zero:
\[ PE_{ground} = 0 \, \text{eV} \]

### Step 3: Calculate the potential energy in the first excited state
For the first excited state (\( n = 2 \)):
\[ PE_{first \, excited} = -\frac{2 \times 13.6 \, \text{eV}}{2^2} = -\frac{2 \times 13.6 \, \text{eV}}{4} = -\frac{27.2 \, \text{eV}}{4} = -6.8 \, \text{eV} \]

### Step 4: Conclusion
Thus, the potential energy of the hydrogen atom in the first excited state is:
\[ PE_{first \, excited} = -6.8 \, \text{eV} \]

### Summary
If the potential energy of a hydrogen atom in the ground state is zero, then its potential energy in the first excited state will be \(-6.8 \, \text{eV}\).

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