The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will be

Options

A. `-6.8 eV`
B. `-13.6 eV`
C. `-27.2 eV`
D. `-54.4 eV`

Answer

To find the energy of a He\(^+\) ion in the first excited state, we can use the formula derived from Bohr's model of the atom. The energy of an electron in a hydrogen-like atom is given by:

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

where:
- \(E_n\) is the energy of the electron in the nth energy level,
- \(Z\) is the atomic number,
- \(n\) is the principal quantum number.

### Step 1: Identify the values for He\(^+\)
For the He\(^+\) ion:
- The atomic number \(Z = 2\) (since helium has 2 protons).
- We are looking for the energy in the first excited state, which corresponds to \(n = 2\).

### Step 2: Substitute the values into the formula
Now, we substitute \(Z = 2\) and \(n = 2\) into the energy formula:

\[ E_2 = -\frac{13.6 \times (2^2)}{(2^2)} \]

### Step 3: Calculate the energy
Calculating the values:

\[ E_2 = -\frac{13.6 \times 4}{4} = -13.6 \, \text{eV} \]

### Conclusion
Thus, the energy of the He\(^+\) ion in the first excited state is:

\[ E_2 = -13.6 \, \text{eV} \]

### Final Answer
The energy of a He\(^+\) ion in the first excited state is \(-13.6 \, \text{eV}\).

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