Find the radius and energy of `He^(+)` ion in the states (a) `n=2, (b) n=3`

Explanation

To find the radius and energy of the `He^(+)` ion in the states \( n=2 \) and \( n=3 \), we can use the formulas for the radius and energy of hydrogen-like atoms.

### Step-by-Step Solution:

### Part (a): For \( n=2 \)

1. **Calculate the Radius \( R \)**:
The formula for the radius of a hydrogen-like ion is given by:
\[ R = \frac{0.53 \, n^2}{Z} \, \text{(in angstroms)} \]
where \( Z \) is the atomic number. For helium ion \( He^+ \), \( Z = 2 \).

Substituting \( n = 2 \) and \( Z = 2 \):
\[ R = \frac{0.53 \times 2^2}{2} = \frac{0.53 \times 4}{2} = \frac{2.12}{2} = 1.06 \, \text{angstroms} \]

2. **Calculate the Energy \( E \)**:
The formula for the energy of a hydrogen-like ion is:
\[ E = -\frac{13.6 \, Z^2}{n^2} \, \text{(in electron volts)} \]
Substituting \( Z = 2 \) and \( n = 2 \):
\[ E = -\frac{13.6 \times 2^2}{2^2} = -\frac{13.6 \times 4}{4} = -13.6 \, \text{eV} \]

### Part (b): For \( n=3 \)

1. **Calculate the Radius \( R \)**:
Using the same formula for the radius:
\[ R = \frac{0.53 \, n^2}{Z} \]
Substituting \( n = 3 \) and \( Z = 2 \):
\[ R = \frac{0.53 \times 3^2}{2} = \frac{0.53 \times 9}{2} = \frac{4.77}{2} = 2.385 \, \text{angstroms} \]

2. **Calculate the Energy \( E \)**:
Using the energy formula:
\[ E = -\frac{13.6 \, Z^2}{n^2} \]
Substituting \( Z = 2 \) and \( n = 3 \):
\[ E = -\frac{13.6 \times 2^2}{3^2} = -\frac{13.6 \times 4}{9} = -\frac{54.4}{9} \approx -6.044 \, \text{eV} \]

### Final Results:
- For \( n=2 \):
- Radius \( R = 1.06 \, \text{angstroms} \)
- Energy \( E = -13.6 \, \text{eV} \)

- For \( n=3 \):
- Radius \( R = 2.385 \, \text{angstroms} \)
- Energy \( E \approx -6.044 \, \text{eV} \)