Excited atoms emit radiations consisting of only certain discrete frequencise or wavelengths. In spectroscopy it is often more convenient to use freuquencies and wave numbers are proportional ot energy and spectroscopy involves transitions between different energy levels . The line spectrum shown by a single electron excited atom (a finger print of an atom) can be given as
`1/(lambda)=barvR_(H).Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
where Z is atomic number of single electron atom and `n_(1),n_(2)` are integers and if `n_(2)gt n_(1),` then emission spectrum is noticed and if `n_(2)ltn_(1),` then absorption spectrum is noticed.
Every line in spectrum can be represented as a difference of two terms `(R_(H)Z^(2))/(n_(1)^(2)) "and"(R_(H)Z^(2))/(n_(2)^(2)).`
The ratio of wavelength for II line fo Balmer series and I line of Lyman series for H-like species is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem regarding the ratio of wavelengths for the second line of the Balmer series and the first line of the Lyman series for hydrogen-like species, we will follow these steps:

### Step 1: Understand the Series
- **Balmer Series**: Transitions that end at \( n = 2 \). The second line corresponds to the transition from \( n = 4 \) to \( n = 2 \).
- **Lyman Series**: Transitions that end at \( n = 1 \). The first line corresponds to the transition from \( n = 2 \) to \( n = 1 \).

### Step 2: Write the Formula
The formula for the wavelength in terms of the energy levels is given by:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R_H \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers.

### Step 3: Calculate Wavelength for Balmer Series
For the second line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Substituting these values into the formula:
\[ \frac{1}{\lambda_B} = R_H Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Calculating the terms:
\[ \frac{1}{\lambda_B} = R_H Z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R_H Z^2 \left( \frac{4 - 1}{16} \right) = R_H Z^2 \left( \frac{3}{16} \right) \]

### Step 4: Calculate Wavelength for Lyman Series
For the first line of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the formula:
\[ \frac{1}{\lambda_L} = R_H Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Calculating the terms:
\[ \frac{1}{\lambda_L} = R_H Z^2 \left( 1 - \frac{1}{4} \right) = R_H Z^2 \left( \frac{3}{4} \right) \]

### Step 5: Find the Ratio of Wavelengths
Now, we need to find the ratio of the wavelengths:
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{1}{\frac{1}{\lambda_L}}}{\frac{1}{\frac{1}{\lambda_B}}} = \frac{R_H Z^2 \left( \frac{3}{4} \right)}{R_H Z^2 \left( \frac{3}{16} \right)} \]
The \( R_H Z^2 \) cancels out:
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{3}{4}}{\frac{3}{16}} = \frac{3}{4} \times \frac{16}{3} = 4 \]

### Conclusion
Thus, the ratio of the wavelengths for the second line of the Balmer series to the first line of the Lyman series is:
\[ \frac{\lambda_B}{\lambda_L} = 4 \]