What is the wave number of 4th line in Balmer series of hydrogen spectrum ?
`R = 1, 109, 677 cm^(-1)`

Options

A. `26630 cm^(-1)`
B. `24360 cm^(-1)`
C. `24730 cm^(-1)`
D. `24372 cm^(-1)`

Explanation

To find the wave number of the 4th line in the Balmer series of the hydrogen spectrum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to electron transitions from higher energy levels (n ≥ 3) to n = 2. The lines in the Balmer series are labeled based on the value of n2 (the higher energy level), where n1 is always 2.

2. **Identify the Transition for the 4th Line**:
For the 4th line in the Balmer series:
- The transitions are:
- 1st line: n2 = 3
- 2nd line: n2 = 4
- 3rd line: n2 = 5
- 4th line: n2 = 6
Therefore, for the 4th line, we have n1 = 2 and n2 = 6.

3. **Use the Formula for Wave Number**:
The formula for calculating the wave number (ν̅) is given by:
\[ \nu̅ = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where R is the Rydberg constant.

4. **Substitute the Values**:
Given:
- R = 1,097,677 cm⁻¹
- n1 = 2
- n2 = 6

Substitute these values into the formula:
\[ \nu̅ = 1,097,677 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \]

5. **Calculate the Terms**:
- Calculate \( \frac{1}{2^2} = \frac{1}{4} = 0.25 \)
- Calculate \( \frac{1}{6^2} = \frac{1}{36} \approx 0.02778 \)

Now, subtract these values:
\[ 0.25 - 0.02778 = 0.22222 \]

6. **Final Calculation**:
Now, multiply by R:
\[ \nu̅ = 1,097,677 \times 0.22222 \approx 24372.2 \text{ cm}^{-1} \]

7. **Final Result**:
The wave number of the 4th line in the Balmer series of hydrogen spectrum is approximately:
\[ \nu̅ \approx 24372 \text{ cm}^{-1} \]