The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom is

Options

A. `1215 Å`
B. `1640 Å`
C. `2430 Å`
D. `4687 Å`

Explanation

To find the wavelength of the second spectral line in the Balmer series of singly ionized helium (He\(^+\)), we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 2) to the n = 2 level. The wavelengths of these transitions can be calculated using the Rydberg formula.

### Step 2: Write the Rydberg Formula
The Rydberg formula for the wavelength of spectral lines is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (1 for hydrogen, 2 for singly ionized helium),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 3: Calculate for Hydrogen
For the first spectral line of the Balmer series in hydrogen (H), the transition is from \( n_2 = 3 \) to \( n_1 = 2 \). Given that the wavelength is \( 6561 \, \text{Å} \):

\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda_H} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36} \]

### Step 4: Set Up the Equation
Given \( \lambda_H = 6561 \, \text{Å} = 6561 \times 10^{-10} \, \text{m} \):

\[ \frac{1}{6561 \times 10^{-10}} = R \cdot \frac{5}{36} \]

This is our first equation.

### Step 5: Calculate for Singly Ionized Helium
For the second spectral line in the Balmer series of singly ionized helium (He\(^+\)), the transition is from \( n_2 = 4 \) to \( n_1 = 2 \):

\[ \frac{1}{\lambda_{He}} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda_{He}} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = R \cdot 4 \left( \frac{4 - 1}{16} \right) = R \cdot 4 \cdot \frac{3}{16} = R \cdot \frac{3}{4} \]

### Step 6: Set Up the Second Equation
Now we have:

\[ \frac{1}{\lambda_{He}} = R \cdot \frac{3}{4} \]

### Step 7: Relate the Two Equations
Now we can relate \( \lambda_{He} \) to \( \lambda_H \):

\[ \frac{\lambda_{He}}{\lambda_H} = \frac{R \cdot \frac{4}{3}}{R \cdot \frac{5}{36}} = \frac{4}{3} \cdot \frac{36}{5} = \frac{144}{15} = \frac{48}{5} \]

### Step 8: Solve for \( \lambda_{He} \)
Now substituting \( \lambda_H = 6561 \times 10^{-10} \):

\[ \lambda_{He} = \frac{48}{5} \cdot 6561 \times 10^{-10} \]

Calculating this gives:

\[ \lambda_{He} = 1215 \times 10^{-10} \, \text{m} = 1215 \, \text{Å} \]

### Conclusion
The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is \( 1215 \, \text{Å} \).

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