The shortest and longest wave number respectively in H spectrum of Lyman series is : (R = Rydberg constant)

Options

A. (1) `(3)/(4)R,R`
B. (2) `(1)/(R ),(4)/(3)R`
C. (3) `R,(4)/(3) R`
D. (4) `R, (3)/(4) R`

Explanation

To find the shortest and longest wave numbers in the hydrogen spectrum of the Lyman series, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electronic transitions in hydrogen where the electron falls to the first energy level (n1 = 1). The second energy level (n2) can be any level greater than 1 (n2 = 2, 3, 4, ...).

### Step 2: Use the Rydberg Formula
The wave number (ν̅) can be calculated using the Rydberg formula:
\[ \nu̅ = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant,
- \( n_1 \) is the lower energy level (for Lyman series, \( n_1 = 1 \)),
- \( n_2 \) is the higher energy level (can be 2, 3, 4, ...).

### Step 3: Calculate the Shortest Wave Number
For the shortest wave number, we take the transition from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ \nu̅_{shortest} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Substituting the values:
\[ \nu̅_{shortest} = R_H \left( 1 - \frac{1}{4} \right) \]
\[ \nu̅_{shortest} = R_H \left( \frac{3}{4} \right) \]
Thus, the shortest wave number is:
\[ \nu̅_{shortest} = \frac{3}{4} R_H \]

### Step 4: Calculate the Longest Wave Number
For the longest wave number, we take the transition from \( n_2 = \infty \) to \( n_1 = 1 \):
\[ \nu̅_{longest} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \nu̅_{longest} = R_H \left( 1 - 0 \right) \]
Thus, the longest wave number is:
\[ \nu̅_{longest} = R_H \]

### Final Answer
The shortest and longest wave numbers in the hydrogen spectrum of the Lyman series are:
- Shortest wave number: \( \frac{3}{4} R_H \)
- Longest wave number: \( R_H \)

### Summary
The answer to the question is:
- Shortest wave number: \( \frac{3}{4} R_H \)
- Longest wave number: \( R_H \)