Calculate Wavelength of Second Balmer Line in Singly Ionized Helium

To determine the wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom ($\text{He}^+$), we use the Rydberg formula for 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 atom,
• $n_1$ and $n_2$ are the principal quantum numbers of the lower and higher energy levels, respectively.

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Step 1: First spectral line in the Balmer series of the Hydrogen atom ($H$)

For hydrogen, the atomic number is $Z_{\text{H}} = 1$. The Balmer series corresponds to transitions ending at $n_1 = 2$. The first line of the Balmer series corresponds to a transition from $n_2 = 3$ to $n_1 = 2$.

Given:

• $\lambda_{\text{H}, 1} = 6561 \text{ \AA}$

Applying the Rydberg formula:
$\frac{1}{\lambda_{\text{H}, 1}} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$

$\frac{1}{6561} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$

Solving for $R$:
$R = \frac{36}{5 \times 6561} \text{ \AA}^{-1}$

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Step 2: Second spectral line in the Balmer series of singly-ionized Helium ($\text{He}^+$)

For singly-ionized helium, the atomic number is $Z_{\text{He}} = 2$. The second line in the Balmer series corresponds to a transition from $n_2 = 4$ to $n_1 = 2$.

Applying the Rydberg formula:
$\frac{1}{\lambda_{\text{He}, 2}} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$

$\frac{1}{\lambda_{\text{He}, 2}} = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$

Taking the reciprocal gives:
$\lambda_{\text{He}, 2} = \frac{4}{3R}$

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Step 3: Calculation

Substitute the value of $R$ into the expression for $\lambda_{\text{He}, 2}$:

$\lambda_{\text{He}, 2} = \frac{4}{3 \times \left( \frac{36}{5 \times 6561} \right)} = \frac{4 \times 5 \times 6561}{3 \times 36}$

$\lambda_{\text{He}, 2} = \frac{20 \times 6561}{108}$

Simplifying the fraction:
$\frac{6561}{108} = \frac{243}{4} = 60.75$

$\lambda_{\text{He}, 2} = 20 \times 60.75 = 1215 \text{ \AA}$

Thus, the wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is $1215\text{ \AA}$.

Correct Answer: A