Upon transition to the ground state, the atom emits a photon with a wavelength of 0.43 μm. Find the energy of a quantum.

To find out the energy of the quantum under consideration, we will use the formula: Ekv = h * C / λ.

Const: h – constant of the quantum of action, Planck (h = 6.626 * 10 ^ -34 J * s); C – the speed of propagation of light (C ≈ 3 * 10 ^ 8 m / s).

Data: λ – known photon wavelength (λ = 0.43 μm = 0.43 * 10 ^ -6 m).

Let’s perform the calculation: Ekv = h * C / λ = 6.62 ^ 6 * 10 ^ -34 * 3 * 10 ^ 8 / (0.43 * 10 ^ -6) ≈ 462.3 * 10 ^ -20 J.

Answer: The energy of the quantum under consideration should be 462.3 zJ.