To calculate the wavelength of a Pfund line corresponding to n₂ = 6, we can use the Rydberg formula for the hydrogen atom. The Rydberg formula gives the wavelength of spectral lines emitted or absorbed by hydrogen atoms in the Balmer series, which includes the Pfund series. The formula is as follows:
1/λ = R_H * (1/n₁² - 1/n₂²)
where λ is the wavelength of the spectral line, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10⁷ m⁻¹), and n₁ and n₂ are the principal quantum numbers of the initial and final energy levels, respectively.
In the case of the Pfund series, n₁ = 5 (since it corresponds to transitions from higher energy levels to the n = 5 energy level) and n₂ = 6. Plugging these values into the Rydberg formula, we get:
1/λ = 1.097 × 10⁷ * (1/5² - 1/6²)
Simplifying the equation:
1/λ = 1.097 × 10⁷ * (1/25 - 1/36)
1/λ = 1.097 × 10⁷ * (36/900 - 25/900)
1/λ = 1.097 × 10⁷ * (11/900)
1/λ = 1.2122 × 10⁴ m⁻¹
Taking the reciprocal of both sides:
λ = 8.24 × 10⁻⁵ m
So, the wavelength of a Pfund line corresponding to n₂ = 6 is approximately 8.24 × 10⁻⁵ meters (82.4 micrometers).