Radio astronomers often study the radiation emitted by a hydrogen atom from a transition between the two hyperfine levels associated with the ground state.This radiation has a wavelength of 21 cm.What is the energy difference between the hyperfine levels? (1 eV = 1.60 × 10^-19 J)

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Quizplus · Accepted Answer

The energy difference can be calculated using the formula E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J·s), c is the speed of light (3 × 10^8 m/s), and λ is the wavelength (21 cm = 0.21 m). Converting the result to eV gives approximately 5.9 × 10^-6 eV.

Radio Astronomers Often Study the Radiation Emitted by a Hydrogen