The energy associatied with the first orbit in the hydrogen atom is ` - 2.17 xx 10^(-18) "J atom" ^(-1)` . What is the energy associated with the fifth orbit ?

Explanation

To find the energy associated with the fifth orbit of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The energy associated with the first orbit (n=1) in the hydrogen atom is given as:
\[ E_1 = -2.17 \times 10^{-18} \text{ J atom}^{-1} \]

2. **Use the Energy Formula:**
- According to Bohr's model, the energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
- For hydrogen, the atomic number \(Z = 1\).

3. **Convert the Energy Formula to Joules:**
- We need to convert the energy from electron volts (eV) to joules (J). We know that:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
- Therefore, the formula in joules becomes:
\[ E_n = -\frac{13.6 \times Z^2}{n^2} \times 1.602 \times 10^{-19} \text{ J} \]

4. **Plug in the Values for the Fifth Orbit (n=5):**
- Substitute \(Z = 1\) and \(n = 5\) into the formula:
\[ E_5 = -\frac{13.6 \times 1^2}{5^2} \times 1.602 \times 10^{-19} \]

5. **Calculate \(E_5\):**
- First, calculate \(5^2\):
\[ 5^2 = 25 \]
- Now substitute this back into the equation:
\[ E_5 = -\frac{13.6}{25} \times 1.602 \times 10^{-19} \]
- Calculate \(-\frac{13.6}{25}\):
\[ -\frac{13.6}{25} = -0.544 \]
- Now multiply by \(1.602 \times 10^{-19}\):
\[ E_5 = -0.544 \times 1.602 \times 10^{-19} \approx -8.69 \times 10^{-20} \text{ J} \]

6. **Final Answer:**
- The energy associated with the fifth orbit of the hydrogen atom is:
\[ E_5 \approx -8.68 \times 10^{-20} \text{ J atom}^{-1} \]