Final answer:
The wavelength of light emitted by an atom as it relaxes back to the ground state with energy 3.06 × 10-19 J can be calculated using the Planck-Einstein relation. Upon solving, it is found that the wavelength is approximately 647.7 nm, indicating emitted light in the red part of the visible spectrum.
Step-by-step explanation:
To find the wavelength of light emitted by an atom upon relaxation back to the ground state, we can employ the Planck-Einstein relation which equates the energy difference of the atom's two energy states with the energy of the emitted photon. We are given that the atom has energy 3.06 × 10-19 J above the ground state, and this energy must be equal to the energy of the photon, which can be described using the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 × 10-34 J·s), c is the speed of light (3.00 × 108 m/s), and λ is the wavelength of the photon.
The expression can be rearranged to solve for λ:
λ = hc/E
Substitute in the known values: λ = (6.626 × 10-34 J·s × 3.00 × 108 m/s) / (3.06 × 10-19 J)
Upon calculation, λ equals approximately 647.7 nm, which means the wavelength of the light emitted is in the visible spectrum, specifically a red color.