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, follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The shortest wavelength transition occurs when an electron transitions from the highest energy level (n = ∞) to n = 2.

### Step 2: Identify the Energy Levels
For the shortest wavelength transition in the Balmer series:
- Initial energy level (n2) = ∞ (infinity)
- Final energy level (n1) = 2

### Step 3: 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 \( 109678 \, \text{cm}^{-1} \)
- \( n_1 = 2 \)
- \( n_2 = \infty \)

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

### Step 5: Calculate the Wave Number
Now, substituting the value of \( R_H \):
\[ \nu̅ = 109678 \cdot \frac{1}{4} = \frac{109678}{4} = 27419.5 \, \text{cm}^{-1} \]

### Final Answer
The wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen is:
\[ \nu̅ = 27419.5 \, \text{cm}^{-1} \]

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