The wavelength of the first line of the Balmer series of hydrogen atom is `lamda`. The wavelength of the corresponding line line of doubly ionized lithium atom is

Options

A. `(lamda)/(3)`
B. `(lamda)/(4)`
C. `(lamda)/(9)`
D. `(lamda)/(27)`

Explanation

To find the wavelength of the corresponding line of the doubly ionized lithium atom (Li²⁺) in relation to the first line of the Balmer series of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Balmer Series for Hydrogen
The Balmer series corresponds to transitions where the electron falls to the second energy level (n=2) from higher energy levels (n=3, 4, ...). The first line of the Balmer series occurs when the electron transitions from n=3 to n=2.

### Step 2: Write the Formula for Wavelength
The wavelength of light emitted during a transition can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 3: Calculate Wavelength for Hydrogen
For the first line of the Balmer series in hydrogen:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the right-hand side:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda} = R \cdot \frac{5}{36} \]
This gives us:
\[ \lambda = \frac{36}{5R} \]

### Step 4: Calculate Wavelength for Doubly Ionized Lithium (Li²⁺)
Doubly ionized lithium (Li²⁺) behaves like a hydrogen atom but with a nuclear charge \( Z = 3 \). The Rydberg formula for Li²⁺ will be:
\[ \frac{1}{\lambda'} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For Li²⁺, we again have:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting \( Z = 3 \):
\[ \frac{1}{\lambda'} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \cdot 9 \left( \frac{1}{4} - \frac{1}{9} \right) \]
We already calculated \( \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \), so:
\[ \frac{1}{\lambda'} = R \cdot 9 \cdot \frac{5}{36} = \frac{45R}{36} = \frac{5R}{4} \]
Thus,
\[ \lambda' = \frac{4}{5R} \]

### Step 5: Relate Wavelengths
We have:
\[ \lambda = \frac{36}{5R} \quad \text{and} \quad \lambda' = \frac{4}{5R} \]
Now, we can express \( \lambda' \) in terms of \( \lambda \):
\[ \lambda' = \frac{4}{5R} = \frac{4}{36} \cdot \lambda = \frac{\lambda}{9} \]

### Final Answer
The wavelength of the corresponding line of the doubly ionized lithium atom is:
\[ \lambda' = \frac{\lambda}{9} \]