Calculate Longest Wavelength of Helium Ion in Balmer Series

To find the longest wavelength of $\text{He}^+$ in the Balmer series, we use the Rydberg formula for hydrogen and hydrogen-like species:

$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where:

• $R$ is the Rydberg constant,
• $Z$ is the atomic number of the species,
• $n_1$ and $n_2$ are the principal quantum numbers of the lower and higher energy levels, respectively.

Step 1: Calculate the Rydberg constant in terms of $x$

For the Hydrogen atom ($Z = 1$), the Lyman series corresponds to transitions to $n_1 = 1$. The shortest wavelength ($\lambda_{\text{min}}$) occurs for the transition from $n_2 = \infty$ to $n_1 = 1$:

$\frac{1}{x} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$

$\frac{1}{x} = R(1 - 0)$

$R = \frac{1}{x}$

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Step 2: Calculate the longest wavelength in the Balmer series for $\text{He}^+$

For the $\text{He}^+$ ion ($Z = 2$), the Balmer series corresponds to transitions to $n_1 = 2$. The longest wavelength ($\lambda_{\text{max}}$) occurs for the minimum energy transition, which is from $n_2 = 3$ to $n_1 = 2$:

$\frac{1}{\lambda_{\text{max}}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$

$\frac{1}{\lambda_{\text{max}}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right)$

$\frac{1}{\lambda_{\text{max}}} = 4R \left( \frac{5}{36} \right) = \frac{5R}{9}$

Substitute $R = \frac{1}{x}$ into the equation:

$\frac{1}{\lambda_{\text{max}}} = \frac{5}{9x}$

$\lambda_{\text{max}} = \frac{9x}{5}$

Thus, the correct option is A.