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. 2 v

Explanation

To solve the problem of finding the frequency of the first line in the Lyman series for doubly ionized Lithium (Li²⁺), we can follow these steps:

### Step 1: Understand the relationship between energy and frequency
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E \propto \frac{Z^2}{n^2} \]
where \(Z\) is the atomic number and \(n\) is the principal quantum number (the energy level).

### Step 2: Relate energy to frequency
The energy can also be expressed in terms of frequency (\(v\)) using Planck's equation:
\[ E = h \cdot v \]
where \(h\) is Planck's constant. Thus, we can say:
\[ h \cdot v \propto \frac{Z^2}{n^2} \]
This implies:
\[ v \propto \frac{Z^2}{n^2} \]

### Step 3: Remove the proportionality
We can express the frequency as:
\[ v = k \cdot \frac{Z^2}{n^2} \]
where \(k\) is a constant.

### Step 4: Determine the frequency for Hydrogen
For Hydrogen (\(Z = 1\)), the frequency of the first line in the Lyman series is given as \(v = \mu\). Substituting \(Z = 1\) into the equation gives:
\[ \mu = k \cdot \frac{1^2}{n^2} = k \cdot \frac{1}{n^2} \]

### Step 5: Determine the frequency for doubly ionized Lithium
For doubly ionized Lithium (\(Li^{2+}\)), the atomic number \(Z = 3\). The frequency for the first line in the Lyman series for Lithium can be expressed as:
\[ v_1 = k \cdot \frac{3^2}{n^2} = k \cdot \frac{9}{n^2} \]

### Step 6: Relate the two frequencies
Since both frequencies correspond to the same principal quantum number \(n\), we can set up the ratio:
\[ \frac{v_1}{\mu} = \frac{k \cdot \frac{9}{n^2}}{k \cdot \frac{1}{n^2}} = \frac{9}{1} \]
This simplifies to:
\[ v_1 = 9 \mu \]

### Conclusion
Thus, the frequency of the corresponding line in the spectrum of doubly ionized Lithium is:
\[ \boxed{9 \mu} \]