The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen is

Options

A. `15.2 xx 10^(6)m^(-1)`
B. `13.6 xx 10^(6)m^(-1)`
C. `1.5 xx 10^(6)m^(-1)`
D. `1.3 xx 10^(6)m^(-1)`

Explanation

To find the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Balmer Series**: The Balmer series corresponds to transitions where the electron falls to the n=2 energy level. The longest wavelength transition occurs when the electron transitions from the nearest higher energy level (n=3) to n=2.

2. **Use the Rydberg Formula**: The wave number (denoted as \( \bar{\nu} \)) can be calculated using the Rydberg formula:
\[ \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).
- \( n_1 \) is the lower energy level (2 for Balmer series).
- \( n_2 \) is the higher energy level (3 for the longest wavelength transition).

3. **Substitute the Values**: Plugging in the values for \( n_1 \) and \( n_2 \):
\[ \bar{\nu} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Calculate the Difference**: Calculate \( \frac{1}{4} - \frac{1}{9} \):
- Find a common denominator (36):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
- Thus,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

5. **Final Calculation**: Now substitute back into the Rydberg formula:
\[ \bar{\nu} = R_H \left( \frac{5}{36} \right) = 1.097 \times 10^7 \, \text{m}^{-1} \times \frac{5}{36} \]
- Calculate:
\[ \bar{\nu} = \frac{5.485 \times 10^7}{36} \approx 1.5 \times 10^6 \, \text{m}^{-1} \]

### Final Answer:
The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen is approximately \( 1.5 \times 10^6 \, \text{m}^{-1} \).

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