Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Explanation

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

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions where an electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5, ...). The longest wavelength transition occurs when the electron falls from the nearest higher energy level, which is n=3.

### Step 2: Identify the Formula for Wave Number
The wave number (ν̅) can be calculated using the formula:
\[ \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant, approximately \( 109677 \, \text{cm}^{-1} \)
- \( n_1 \) is the lower energy level (for Balmer series, \( n_1 = 2 \))
- \( n_2 \) is the higher energy level (for the longest wavelength, \( n_2 = 3 \))

### Step 3: Substitute Values into the Formula
Substituting the values into the formula:
\[ \bar{\nu} = 109677 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \):
\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]
Now, calculate the difference:
\[ 0.25 - 0.1111 = 0.1389 \]

### Step 4: Calculate the Wave Number
Now, substituting this back into the wave number formula:
\[ \bar{\nu} = 109677 \times 0.1389 \]
Calculating this gives:
\[ \bar{\nu} \approx 15232.9 \, \text{cm}^{-1} \]

### Step 5: Final Result
Thus, the wave number for the longest wavelength transition in the Balmer series is:
\[ \bar{\nu} \approx 15232.9 \, \text{cm}^{-1} \]