To solve the problem of calculating the longest wavelength in the Balmer series of singly ionized helium (He⁺), we will follow these steps:
### Step 1: Understand the Formula
The formula given for the wavelength (λ) in the Balmer series is:
\[
\frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right)
\]
Where:
- \( R_H \) is the Rydberg constant for hydrogen, which is \( 109.678 \, \text{cm}^{-1} \).
- \( Z \) is the atomic number of the ion (for He⁺, \( Z = 2 \)).
- \( n \) is the principal quantum number, which takes values \( n = 3, 4, 5, \ldots \) for the Balmer series.
### Step 2: Identify the Longest Wavelength
To find the longest wavelength, we need to minimize the term \( \frac{1}{n^2} \). The smallest value of \( n \) in the Balmer series is \( n = 3 \).
### Step 3: Substitute Values into the Formula
Now we substitute \( Z = 2 \) and \( n = 3 \) into the formula:
\[
\frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
\]
Calculating the terms:
- \( R_H = 109.678 \, \text{cm}^{-1} \)
- \( Z^2 = 2^2 = 4 \)
- \( \frac{1}{2^2} = \frac{1}{4} \)
- \( \frac{1}{3^2} = \frac{1}{9} \)
Now, calculate \( \frac{1}{4} - \frac{1}{9} \):
\[
\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}
\]
### Step 4: Plug in the Values
Now substituting back into the equation:
\[
\frac{1}{\lambda} = 109.678 \cdot 4 \cdot \frac{5}{36}
\]
Calculating this step-by-step:
1. \( 4 \cdot \frac{5}{36} = \frac{20}{36} = \frac{5}{9} \)
2. Now multiply by \( R_H \):
\[
\frac{1}{\lambda} = 109.678 \cdot \frac{5}{9}
\]
3. Calculate \( 109.678 \cdot \frac{5}{9} \):
\[
\frac{1}{\lambda} \approx 60.932 \, \text{cm}^{-1}
\]
### Step 5: Calculate λ
Now, take the reciprocal to find λ:
\[
\lambda = \frac{1}{60.932} \approx 0.0164 \, \text{cm}
\]
### Step 6: Convert to Angstroms
Since \( 1 \, \text{cm} = 10^{10} \, \text{Å} \):
\[
\lambda \approx 0.0164 \, \text{cm} \times 10^{10} \, \text{Å/cm} = 1641.6 \, \text{Å}
\]
### Final Answer
The longest wavelength in the Balmer series of singly ionized helium (He⁺) is approximately:
\[
\lambda \approx 1641.6 \, \text{Å}
\]
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