Understanding the Concept:
According to Rydberg's formula modified for hydrogen-like species, the wave number ($\bar{v}$) of a photon emitted during an electronic transition between an initial higher shell $n_2$ and a final lower shell $n_1$ is given by:
\[
\bar{v} = \frac{1}{\lambda} = R_{\text{H}} \cdot Z^2 \cdot \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)
\]
where $R_{\text{H}} \approx 1.09677 \times 10^7\text{ m}^{-1}$ is the Rydberg constant and $Z$ is the atomic number. The frequency ($v$) of the emitted radiation is calculated using the wave speed relationship:
\[
v = c \cdot \bar{v} = c \cdot R_{\text{H}} \cdot Z^2 \cdot \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)
\]
Step 1: Substitute the given parameters into the emission equation.
For a helium ion ($\text{He}^+$), the atomic number is $Z = 2$. The transition occurs from $n_2 = 5$ to $n_1 = 3$.
Using the combined constant product $c \cdot R_{\text{H}} \approx 3.29 \times 10^{15}\text{ Hz}$:
\[
v = (3.29 \times 10^{15}) \times (2)^2 \times \left(\frac{1}{3^2} - \frac{1}{5^2}\right)
\]
\[
v = (3.29 \times 10^{15}) \times 4 \times \left(\frac{1}{9} - \frac{1}{25}\right)
\]
Step 2: Simplify the fractional expressions and calculate frequency.
Find a common denominator for the terms inside the parentheses:
\[
\frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225}
\]
Now, compute the product value:
\[
v = 3.29 \times 10^{15} \times 4 \times \frac{16}{225} = 3.29 \times 10^{15} \times \frac{64}{225}
\]
\[
v \approx 3.29 \times 10^{15} \times 0.28444 \approx 0.9358 \times 10^{15}\text{ s}^{-1} \approx 9.39 \times 10^{14}\text{ s}^{-1}
\]