Be Ion: Calculating Longest Wavelength Absorbed

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Homework Statement


The element Be has 4 electrons and 4 protons in the atom. An ion of Be has all the electrons removed except 1, so it resembles a hydrogen atom with a nuclear charge of +4 e.

Constants:

E1(Hydrogen)= -13.6eV, r1(Hydrogen)= 5.29x10-11m.

If a gas is made up of ions, as described, in their lowest energy states, and is bombarded with white light, what is the longest wavelength, in nm, which will be absorbed ?


Homework Equations


λ = hc/E


The Attempt at a Solution


Total energy of the lowest energy state = E = -13.6* no. of protons^2 / n^2 = ((-13.6 * 4^2)/1^2) = -217.6
λ = hc/E = (6.63 E –34 * 3E+8) / (-217.6 * 1.602E-19) = -5.7E-9 m = -5.7 nm
the answer is 7.625nm...

White light doesn't have a wavelength...I have no idea to solve this
 
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White light is a mixture of all wavelengths, so all of the necessary wavelengths for the possible transitions are present. The only limiting factor here is that the Be ions will move to certain discrete energy levels. You are given that the atoms are initially in the n=1 state, so the possible transitions are from n=1 to n=2, n=1 to n=3, n=1 to n=4...etc. For each of these transitions, the energy absorbed by the ion will have a different value. For an absorption of a given amount of energy (say \Delta E), what wavelength of light is needed? Do the necessary wavelengths increase or decrease as \Delta E gets larger? What is the smallest possible \Delta E?
 
oh, so n = 1 to n= 2 gives the smallest possible "delta E"
λ = hc/(E1 - E2) = (6.63 E –34 * 3E+8) / (-217.6 + 54.4) * 1.602E-19) = 7.60E-9 = 7.60nm ?
 
Shouldn't it be E2-E1 instead of E1-E2? (You're getting a negative sign that you didn't put next to your final answer)

Other than that, it looks good to me.
 
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