If the wavelength of the first member of Balmer series of hydrogen spectrum is `6564A^(@)` , the wavelength of second member of Balmer series will be:

Options

A. `1215A^(@)`
B. `4848A^(@)`
C. `6050A^(@)`
D. data given is insufficient to calculate the value

Explanation

To find the wavelength of the second member of the Balmer series for the hydrogen spectrum, 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 ≥ 3) to the second energy level (n = 2). The first member of the Balmer series corresponds to the transition from n = 3 to n = 2.

### Step 2: Given Information
We are given the wavelength of the first member of the Balmer series:
- Wavelength (λ₁) = 6564 Å (Angstroms)

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during an electron transition is given by:

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

Where:
- \( R_{\infty} \) = Rydberg constant = \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 \) = lower energy level (for the Balmer series, \( n_1 = 2 \))
- \( n_2 \) = higher energy level (for the first member, \( n_2 = 3 \))

### Step 4: Calculate \( R_{\infty} \)
For the first member (n = 3 to n = 2), we can rearrange the formula to find \( R_{\infty} \):

\[ \frac{1}{\lambda_1} = R_{\infty} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Substituting the values:

\[ \frac{1}{6564 \times 10^{-10}} = R_{\infty} \left( \frac{1}{4} - \frac{1}{9} \right) \]

Calculating the right side:

\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

Now substituting this back into the equation:

\[ \frac{1}{6564 \times 10^{-10}} = R_{\infty} \left( \frac{5}{36} \right) \]

### Step 5: Solve for \( R_{\infty} \)
Now we can solve for \( R_{\infty} \):

\[ R_{\infty} = \frac{1}{6564 \times 10^{-10}} \times \frac{36}{5} \]

### Step 6: Calculate Wavelength of the Second Member
Now we need to find the wavelength of the second member of the Balmer series, which corresponds to the transition from n = 4 to n = 2. Using the Rydberg formula again:

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

Calculating the right side:

\[ \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \]

Now substituting back into the equation:

\[ \frac{1}{\lambda_2} = R_{\infty} \left( \frac{3}{16} \right) \]

### Step 7: Substitute \( R_{\infty} \) and Solve for \( \lambda_2 \)
We can substitute the value of \( R_{\infty} \) that we calculated earlier and solve for \( \lambda_2 \).

### Final Calculation
After calculating, we will find that:

\[ \lambda_2 \approx 4862 \, \text{Å} \]

### Conclusion
Thus, the wavelength of the second member of the Balmer series is approximately 4862 Å.

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