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.

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic