If the wave number of `1^(st)` line of Balmer series of H-atom is 'x' then :

Options

A. Wave number of `1^(st)` line of lyman series of the `He^(+)` ion will be `(108 x)/5`
B. Wave number of `1^(st)` line of lyman series of the `He^(+)` ion will be `(36x)/5`
C. The wave length of `2^(nd)` line of lyman series of H-atom is `5/(32x)`
D. The wave length of `2^(nd)` line of lyman series of H-atom is `(32x)/5`

Explanation

To solve the problem regarding the wave number of the first line of the Balmer series of the hydrogen atom, we will follow these steps:

### Step 1: Understand the Formula
The wave number \( \bar{v} \) (in cm\(^{-1}\)) for the transition between energy levels in a hydrogen atom can be calculated using the formula:
\[ \bar{v} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 2: Identify the Transition for the First Line of the Balmer Series
For the first line of the Balmer series, the transition occurs from \( n_2 = 3 \) to \( n_1 = 2 \).

### Step 3: Calculate the Wave Number
Substituting the values into the formula:
\[ \bar{v} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus, we have:
\[ \bar{v} = R \cdot \frac{5}{36} \]

### Step 4: Relate to Given Wave Number \( x \)
We are given that the wave number of the first line of the Balmer series is \( x \). Therefore, we can express \( x \) as:
\[ x = R \cdot \frac{5}{36} \]

### Step 5: Determine the Wave Number for the First Line of the Lyman Series
For the first line of the Lyman series, the transition is from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ \bar{v}_{Lyman} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Now, we can express this in terms of \( x \):
\[ \bar{v}_{Lyman} = R \cdot \frac{3}{4} = \frac{3}{4} \cdot \frac{36}{5} x = \frac{27}{5} x \]

### Step 6: Determine the Wavelength of the Second Line of the Lyman Series
The second line of the Lyman series corresponds to a transition from \( n_2 = 3 \) to \( n_1 = 1 \):
\[ \bar{v}_{Lyman, 2} = R \left( 1 - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \cdot \frac{8}{9} \]
Expressing this in terms of \( x \):
\[ \bar{v}_{Lyman, 2} = R \cdot \frac{8}{9} = \frac{8}{9} \cdot \frac{36}{5} x = \frac{64}{45} x \]

### Step 7: Find the Wavelength
The wavelength \( \lambda \) is the inverse of the wave number:
\[ \lambda = \frac{1}{\bar{v}} = \frac{1}{\frac{64}{45} x} = \frac{45}{64 x} \]

### Conclusion
Based on the calculations, we can summarize the results:
- The wave number of the first line of the Lyman series of helium plus is \( \frac{108}{5} x \).
- The wavelength of the second line of the Lyman series of hydrogen atom is \( \frac{45}{64} x \).