When an electron transitions from a higher energy level to a lower energy level in the hydrogen atom, it emits electromagnetic radiation in the form of a photon. The energy of the photon is equal to the difference in energy between the two energy levels.



The energy levels of the hydrogen atom are quantized, meaning that they are discrete and can only take on certain values. The energy of an electron in the hydrogen atom is given by the equation:

E = -13.6/n^2 eV

Where E is the energy, n is the principal quantum number, and eV is electron volts. When an electron transitions from a higher energy level to a lower energy level, it releases energy in the form of a photon. The energy of the photon is given by the equation:

E = hf

Where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon.

The energy difference (ΔE) between the n=5 and n=4 energy levels of the hydrogen atom can be calculated as follows:

ΔE = E5 - E4
ΔE = (-13.6/5^2) - (-13.6/4^2)
ΔE = -1.51 eV



The relationship between the energy and the wavelength of a photon is given by the equation:

E = hc/λ

Where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

As the energy of the photon decreases, its wavelength increases. Therefore, the photon with the highest wavelength will be the one with the lowest energy.



In the case of the hydrogen atom, the transition of the electron from n=5 to n=4 results in the emission of a photon with the highest wavelength because the energy difference between these two energy levels is the smallest, resulting in the lowest energy and therefore the highest wavelength photon.