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

Explanation

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

### Step 1: Identify the Transition
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 shortest wavelength transition occurs when the electron transitions from n=∞ (infinity) to n=2.

### Step 2: Use the Formula for Wave Number
The wave number (ν̅) is given by the formula:
\[ \nu̅ = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 109678 \, \text{cm}^{-1} \).
- \( n_1 \) is the lower energy level (2 for the Balmer series).
- \( n_2 \) is the higher energy level (∞ for the shortest wavelength transition).

### Step 3: Substitute Values into the Formula
Substituting \( n_1 = 2 \) and \( n_2 = \infty \) into the formula:
\[ \nu̅ = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \nu̅ = R_H \left( \frac{1}{4} - 0 \right) = R_H \cdot \frac{1}{4} \]

### Step 4: Calculate the Wave Number
Now, substituting the value of \( R_H \):
\[ \nu̅ = 109678 \cdot \frac{1}{4} = \frac{109678}{4} \approx 27419.5 \, \text{cm}^{-1} \]

### Step 5: Final Result
Thus, the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen is approximately:
\[ \nu̅ \approx 27419.5 \, \text{cm}^{-1} \]