The shortest wavelength radiation in the Balmer series corresponds to the transition from the n = ∞ energy level to the n = 2 energy level. Here's why:

* Balmer Series: This series of spectral lines in the hydrogen atom is characterized by electron transitions that end in the n = 2 energy level.

* Energy Levels: The higher the energy level (n), the further the electron is from the nucleus.

* Wavelength and Energy: A transition from a higher energy level to a lower one releases energy, and this energy is emitted as a photon. The higher the energy difference, the shorter the wavelength of the emitted photon.

Calculation

You can use the Rydberg formula to calculate the wavelength:

1/λ = R(1/2² - 1/n²)

Where:

* λ is the wavelength

* R is the Rydberg constant (approximately 1.097 x 10⁷ m⁻¹)

* n is the initial energy level (in this case, approaching infinity)

As n approaches infinity, the term 1/n² approaches zero. Therefore, the equation simplifies to:

1/λ = R/4

Solving for λ, we get:

λ = 4/R

Plugging in the value for R:

λ ≈ 364.6 nm

Therefore, the shortest wavelength radiation in the Balmer series is approximately 364.6 nanometers. This corresponds to the ultraviolet region of the electromagnetic spectrum.