Q/ An electron in a hydrogen atom undergoes the transitiondepicted to the right. Calculate the wavelength (in nm),frequency (in Hz), and energy (in J) of the photon emitteddue to the relaxation of the electron. Report each value tothree significant figures. Show all work. The image displays a simplified model of an atom, illustrating the Bohr atomic model. At the center, there is a small circle representing the nucleus. Surrounding the nucleus are several concentric circles that represent the electron orbits or energy levels.- The nucleus is depicted as a small dot in the center of the diagram.- There are five concentric circles around the nucleus representing different electron energy levels or shells.- An electron, labeled as "e-", is shown on one of the outer circles. This indicates the electron's position in the energy level.This model is used to describe how electrons are arranged around the nucleus, emphasizing quantized energy levels.

Rydberg's equation will allow to calculate wavelength of the photon emitted by the electron during transition. The equation used to calculate wavelength is, 1λ = R.1nfinal2-1ninitial2 Where, λ = wavelength of the emitted photon.R = Rydberg's constant - 1.0974×107 m -1nfinal = the final energy level ninitial = the initial energy level Energy of of photon is calculated using the formula E = hν = hcλWhere,E = energy,h = Planck constant =6.626 ×10-34 J.sν = frequency = cλc = speed of light = 3.0 ×108 m/sλ = wavelength of photonGiven figure for the transition is, In the given atomic model energy levels are marked. For the figure it is depicted that the electron transition is from n=4 to n=2. That is, nfinal = 2 and ninitial = 4 Wavelength associated with the given transition is calculated as, 1λ = R.1nfinal2-1ninitial21λ =1.0974 ×1071(2)2-1(4)21λ =1.0974 ×107(0.1875)1λ = 2057625 m-1λ=4.86×10-6 m Question ask the wavelength value in the unit nm. Conversion factor 1 m = 109 nm λ = 4.86×10-6 m ×109 nm1 m= 4.86×103 nm Frequency of photon is calculated as, ν = cλ=3.00 ×108 m/s4.86 ×10-6 m=6.17×1013 s-1 (Hz) Energy of photon is calculated as, E = hν =6.626 ×10-34 J.s ×6.17 ×1013s-1=4.09×10-20 J Answer: Wavelength of photon = 4.86*103 nm Frequency of photon = 6.17 * 1013 Hz Energy of photon = 4.09 * 10-20 J