Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when it's electron drops from the n=5 to the n=3 state

IF WE TAKE N=1 IT IS CALLED GROUND STATE. THEN THE OTHER FOLLOWING HIGHER STATES ARE CALLED EXCITED STATES. IF THE ELECTRON IN AN ATOM JUMPS FROM A STATE TO A LOWER STATE, IT LOSES ENERGY. FROM THE GIVEN STATEMENT, THE WAY TO FIND THE ENERGY RELEASED IS GIVEN BY THE FORMULA, E(n)=(-13.6 eV)/n^2. FIRST TO FIND E(5)=(-13.6 eV)/(5)^2, WE GET E(5)=-0.544 eV. E(3)=(-13.6 eV)/(3)^2, WE GET E(3)=-1.5111 eV. THEN WE HAVE TO FIND THE ENERGY TRANSITION LEVEL. ON SUBTRACTING WE GET 0.967eV. THIS ENERGY HAS TO BE CONVERTED IN JOULES. SO WE MAKE E=0.967*(1.60*10^(-19)) J/eV, WHICH CORRESPONDS TO 0.15472*10^(-18) J. WE NEED TO FIND TO THE WAVELENGTH. THE CORRESPONDING FORMULA E = hf = hc/λ, λ = hc/E. BY SUBSTITUTING THE KNOWN VALUES, WE GET THE ANSWER TO BE 1285.548 NM.