The frequency of the first line in Lyman series in the hydrogen spectrum is v. What is the frequency of the corresponding line in the spectrum of doubly ionized Lithium?

Options

A. `v`
B. `3 v`
C. `9 v`
D. `27 v`

Answer

To solve the problem, we need to find the frequency of the first line in the Lyman series for doubly ionized lithium (Li²⁺) and compare it to the frequency of the first line in the Lyman series for hydrogen.

### Step-by-Step Solution:

1. **Understanding the Lyman Series**:
The Lyman series corresponds to transitions from higher energy levels (n ≥ 2) to the ground state (n = 1) in hydrogen and hydrogen-like ions. The frequency of the emitted photon during these transitions can be calculated using the Rydberg formula.

2. **Rydberg Formula for Hydrogen**:
The frequency of the transition for hydrogen can be expressed as:
\[ \nu_H = R_H \left(1 - \frac{1}{n^2}\right) \]
where \( R_H \) is the Rydberg constant for hydrogen and \( n \) is the principal quantum number of the higher energy level.

3. **Calculating Frequency for Hydrogen**:
For the first line in the Lyman series, we consider the transition from \( n = 2 \) to \( n = 1 \):
\[ \nu_H = R_H \left(1 - \frac{1}{2^2}\right) = R_H \left(1 - \frac{1}{4}\right) = R_H \left(\frac{3}{4}\right) \]
Given that this frequency is denoted as \( v \), we have:
\[ v = R_H \left(\frac{3}{4}\right) \]

4. **Rydberg Formula for Doubly Ionized Lithium (Li²⁺)**:
For doubly ionized lithium, which is a hydrogen-like atom with atomic number \( Z = 3 \), the frequency can be expressed as:
\[ \nu_{Li} = R_{Li} \left(1 - \frac{1}{n^2}\right) \]
where \( R_{Li} = Z^2 R_H = 3^2 R_H = 9 R_H \).

5. **Calculating Frequency for Li²⁺**:
For the first line in the Lyman series for Li²⁺, we again consider the transition from \( n = 2 \) to \( n = 1 \):
\[ \nu_{Li} = 9 R_H \left(1 - \frac{1}{2^2}\right) = 9 R_H \left(1 - \frac{1}{4}\right) = 9 R_H \left(\frac{3}{4}\right) \]

6. **Finding the Ratio of Frequencies**:
Now, we can express the frequency of Li²⁺ in terms of \( v \):
\[ \nu_{Li} = 9 \left(\frac{3}{4}\right) R_H = \frac{27}{4} R_H \]
Since \( v = R_H \left(\frac{3}{4}\right) \), we can rewrite \( \nu_{Li} \) as:
\[ \nu_{Li} = 9v \]

### Final Answer:
The frequency of the corresponding line in the spectrum of doubly ionized lithium is \( 9v \).