Find the radius and energy of a `He^(++)`ion in the states (a) `n = 1 , (b) n = 4 and (c) n= 10` is

Explanation

To find the radius and energy of a `He^(++)` ion in different states (n = 1, n = 4, and n = 10), we can use the formulas derived from Bohr's model of the atom.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We need to calculate the radius and energy of the `He^(++)` ion for three different principal quantum numbers (n = 1, n = 4, n = 10).
- The atomic number (Z) for helium (He) is 2.

2. **Formulas to Use:**
- The radius of the nth orbit is given by:
\[ R_n = \frac{0.53 \, n^2}{Z} \, \text{angstroms} \]
- The energy of the nth orbit is given by:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{electron volts} \]

3. **Calculating for n = 1:**
- **Radius:**
\[ R_1 = \frac{0.53 \times 1^2}{2} = \frac{0.53}{2} = 0.265 \, \text{angstroms} \]
- **Energy:**
\[ E_1 = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \, \text{electron volts} \]

4. **Calculating for n = 4:**
- **Radius:**
\[ R_4 = \frac{0.53 \times 4^2}{2} = \frac{0.53 \times 16}{2} = \frac{8.48}{2} = 4.24 \, \text{angstroms} \]
- **Energy:**
\[ E_4 = -\frac{13.6 \times 2^2}{4^2} = -\frac{13.6 \times 4}{16} = -\frac{54.4}{16} = -3.4 \, \text{electron volts} \]

5. **Calculating for n = 10:**
- **Radius:**
\[ R_{10} = \frac{0.53 \times 10^2}{2} = \frac{0.53 \times 100}{2} = \frac{53}{2} = 26.5 \, \text{angstroms} \]
- **Energy:**
\[ E_{10} = -\frac{13.6 \times 2^2}{10^2} = -\frac{13.6 \times 4}{100} = -\frac{54.4}{100} = -0.544 \, \text{electron volts} \]

### Final Results:
- For n = 1:
- Radius \( R_1 = 0.265 \, \text{angstroms} \)
- Energy \( E_1 = -54.4 \, \text{electron volts} \)

- For n = 4:
- Radius \( R_4 = 4.24 \, \text{angstroms} \)
- Energy \( E_4 = -3.4 \, \text{electron volts} \)

- For n = 10:
- Radius \( R_{10} = 26.5 \, \text{angstroms} \)
- Energy \( E_{10} = -0.544 \, \text{electron volts} \)