Calculate the wavelength of the light emitted when an electron in a hydrogen atom n=5 to n=2

For a hydrogen, or hydrogen-like, atom, the expression for the energy of an electron in the nth energy level, in units of electron-volts, is: En = (-13.6 eV)/n2 If an electron moves from a higher energy level to a lower one, as in going from n=5 to n=2, it is a little like falling from a higher shelf to a lower one.  It loses an amount of energy equal to the difference in energy between the two levels.  That energy is emitted as light. So the light, in this case, will have energy equal to: E5 - E2 = [(-13.6 eV)/52] - [(-13.6 eV)/22] = 13.6((-1/25) - (-1/4)) = 13.6((1/4)-(1/25)) = 13.6(21/100) = 2.86 eV The frequency of light, f, is related to its energy by the expression: E = hf, where h = Planck's constant (= 4.14x10-15 eV·s) Knowing E and h, you can solve for f. Then, the frequency of light and its wavelength, λ, are related by: λ = c/f  where c = speed of light (2.99 x 108 m/s) This lets you solve for λ, once you have solved for f.

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The answer should be vaporization 

Is a logical series of steps