To calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen, we need to understand what the Balmer series represents and how the wave number is related to the wavelength of electromagnetic radiation.

The Balmer series corresponds to electron transitions from higher energy levels (n > 2) down to n = 2 in the hydrogen atom. The wavelength of the emitted photon during such a transition is given by the Rydberg formula, which for hydrogen is:

$$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$$

where:

• $$\lambda$$ is the wavelength of the emitted light.
• $$R$$ is the Rydberg constant for hydrogen, approximately $$1.097 \times 10^7 \, \text{m}^{-1}$$.
• $$n$$ is the principal quantum number of the higher energy level (n > 2).

The wave number $$\bar{\nu}$$ is defined as the reciprocal of the wavelength (in meters), thus:

$$\bar{\nu} = \frac{1}{\lambda}$$

Substituting the Rydberg formula for $$\frac{1}{\lambda}$$ gives us:

$$\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$$

The shortest wavelength transition in the Balmer series corresponds to the transition with the largest energy difference, which occurs when the electron comes from an infinitely high energy level (n = ∞) down to n = 2. Therefore, for the shortest wavelength (and hence highest frequency and largest wave number), set n = ∞ in the equation. As $$n$$ approaches ∞, $$\frac{1}{n^2}$$ approaches 0, and the equation simplifies to:

$$\bar{\nu}_{min} = R \left( \frac{1}{2^2} \right) = R \left( \frac{1}{4} \right) = \frac{1.097 \times 10^7 \, \text{m}^{-1}}{4}$$

Performing the calculation:

$$\bar{\nu}_{min} = 2.743 \times 10^6 \, \text{m}^{-1}$$

Thus, the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen is $$2.743 \times 10^6 \, \text{m}^{-1}$$. This corresponds to the transition from n = ∞ to n = 2.