Step 1: Understanding the Concept:
According to the Bohr model of the atom, the energy of an electron residing in the $n$-th orbit of a hydrogen-like species (a one-electron species like $\text{H}$, $\text{He}^+$, $\text{Li}^{2+}$) is strictly quantized and can be calculated using a specific formula involving the atomic number ($Z$) and the principal quantum number ($n$).
Step 2: Key Formula or Approach:
The energy $E_n$ of the $n$-th orbit in Joules is given by:
\[ E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \text{ J} \]
Step 3: Detailed Explanation:
For the helium ion ($\text{He}^+$):
Atomic number, $Z = 2$
We need the energy for the third orbit, so $n = 3$
Substitute these values directly into the formula:
\[ E_3 = -2.18 \times 10^{-18} \times \left( \frac{2^2}{3^2} \right) \]
\[ E_3 = -2.18 \times 10^{-18} \times \left( \frac{4}{9} \right) \]
\[ E_3 = -2.18 \times 10^{-18} \times 0.4444... \]
\[ E_3 \approx -0.9688 \times 10^{-18} \text{ J} \]
To precisely match the scientific notation format presented in the options, adjust the decimal place:
\[ E_3 \approx -9.688 \times 10^{-19} \text{ J} \]
Rounding to two decimal places, we get $-9.69 \times 10^{-19} \text{ J}$.
Step 4: Final Answer:
The energy associated with the third orbit is $-9.69 \times 10^{-19} \text{ J}$.