If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should be

Options

A. `13122 Å`
B. `3280 Å`
C. `4860 Å`
D. `2187 Å`

Explanation

To find the wavelength of the second line of the Balmer series of hydrogen, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electronic transitions in hydrogen where the final energy level (n_final) is 2. The initial energy levels (n_initial) for the first and second lines are 3 and 4, respectively.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of emitted light during these transitions is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]

Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_{final} \) is the final energy level,
- \( n_{initial} \) is the initial energy level.

### Step 3: Calculate for the First Line
For the first line of the Balmer series:
- \( n_{final} = 2 \)
- \( n_{initial} = 3 \)

Substituting these values into the Rydberg formula:

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda_1} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 4: Calculate for the Second Line
For the second line of the Balmer series:
- \( n_{final} = 2 \)
- \( n_{initial} = 4 \)

Substituting these values into the Rydberg formula:

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

Finding a common denominator (16):

\[ \frac{1}{\lambda_2} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 5: Relate the Two Wavelengths
Now we have:
\[ \frac{1}{\lambda_1} = R \left( \frac{5}{36} \right) \]
\[ \frac{1}{\lambda_2} = R \left( \frac{3}{16} \right) \]

Dividing the two equations gives:

\[ \frac{\lambda_1}{\lambda_2} = \frac{5/36}{3/16} = \frac{5 \times 16}{3 \times 36} = \frac{80}{108} = \frac{20}{27} \]

Thus, we can express \( \lambda_2 \) in terms of \( \lambda_1 \):

\[ \lambda_2 = \frac{20}{27} \lambda_1 \]

### Step 6: Substitute the Given Wavelength
Given that \( \lambda_1 = 6561 \, \text{Å} \):

\[ \lambda_2 = \frac{20}{27} \times 6561 \]

Calculating this gives:

\[ \lambda_2 = \frac{20 \times 6561}{27} = 4860 \, \text{Å} \]

### Final Answer
The wavelength of the second line of the Balmer series should be \( 4860 \, \text{Å} \).

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