For the hydrogen spectrum last line of the lyman series has frequency v1 last line of lyman series of `He^(+)` ions has frequency `V_(2)` and `1^(st)` line of Lyman series of `He^(+)` ions has frequency `v_(3)` then

Options

A. `4 v_(1)=v_(2)+v_(3)`
B. `v_(1)=4v_(2)+v_(3)`
C. `v_(2)=v_(3)-v_(1)`
D. `v_(2)=v_(1)+v_(3)`

Answer

To solve the problem, we need to find the relationships between the frequencies of the last line of the Lyman series for hydrogen and the He\(^+\) ion. We will denote the frequencies as follows:

- \( \nu_1 \): Last line of the Lyman series for hydrogen
- \( \nu_2 \): Last line of the Lyman series for He\(^+\) ion
- \( \nu_3 \): First line of the Lyman series for He\(^+\) ion

### Step 1: Calculate \( \nu_1 \) for Hydrogen
The frequency of the last line of the Lyman series for hydrogen can be calculated using the Rydberg formula:

\[ \nu_1 = R_H \cdot c \cdot \left(1 - \frac{1}{\infty^2}\right) \]

Where:
- \( R_H \) is the Rydberg constant for hydrogen
- \( c \) is the speed of light

Since \( n_1 = 1 \) and \( n_2 = \infty \):

\[ \nu_1 = R_H \cdot c \cdot (1 - 0) = R_H \cdot c \]

### Step 2: Calculate \( \nu_2 \) for He\(^+\)
For the He\(^+\) ion, the last line of the Lyman series can be calculated similarly, but with \( Z = 2 \):

\[ \nu_2 = R_H \cdot Z^2 \cdot c \cdot \left(1 - \frac{1}{\infty^2}\right) \]

Substituting \( Z = 2 \):

\[ \nu_2 = R_H \cdot (2^2) \cdot c \cdot (1 - 0) = 4R_H \cdot c \]

### Step 3: Calculate \( \nu_3 \) for He\(^+\)
For the first line of the Lyman series for He\(^+\), we have \( n_1 = 1 \) and \( n_2 = 2 \):

\[ \nu_3 = R_H \cdot Z^2 \cdot c \cdot \left(1 - \frac{1}{2^2}\right) \]

Substituting \( Z = 2 \):

\[ \nu_3 = R_H \cdot (2^2) \cdot c \cdot \left(1 - \frac{1}{4}\right) = 4R_H \cdot c \cdot \left(\frac{3}{4}\right) = 3R_H \cdot c \]

### Step 4: Establish the relationship between \( \nu_1 \), \( \nu_2 \), and \( \nu_3 \)
Now we have:

- \( \nu_1 = R_H \cdot c \)
- \( \nu_2 = 4R_H \cdot c \)
- \( \nu_3 = 3R_H \cdot c \)

We can express the relationship between these frequencies:

\[ \nu_2 = \nu_1 + \nu_3 \]

Substituting the values:

\[ 4R_H \cdot c = R_H \cdot c + 3R_H \cdot c \]

This simplifies to:

\[ 4R_H \cdot c = 4R_H \cdot c \]

This confirms that the relationship holds true.

### Conclusion
The relationship between the frequencies is:

\[ \nu_2 = \nu_1 + \nu_3 \]

Thus, the correct option is the one that states this relationship.