# Chemistry

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Determine which H atom transition(iinitital, nfinal) was responsible for the absorption of a photon with a frequency of 4.57 x 10^14 Hz. [RH = 2.18 x 10^-18 J; h = 6.63 x 10^-34 J s]

• Chemistry -

(a) First find the transition energy from the frequency by using the Planck equation:
deltaE = (6.63x10^-34Js)(frequency)
(b) Use the equation below:
deltaE = (-2.18x10^-18J)(1/b^2 - 1/a^2)
a and b are small integers with b>a
Calculate deltaE using trial values for b and a and repeat this calculation until the value you get matches the value you got for part (a). Start with b = 3 and a = 2. If it doesn't work try other combinations.

• Chemistry -

E = hf. You have h and frequency. Calculate E (in Joules) for the transition.

I think the easiest way to go about this is to calculate E for the energy levels of the H atom. It's a little work.
E = 2.180 x 10^-18/N^2.
Now substitute 1 for N and calculate E.
Substitute 2 for N and calculate E.
Substitute 3 for N and calculate E.
Do this for N = 4, 5, 6, and 7 (probably that's enough). Now subtract E2-E1, E3-E1, E3-E2, etc. You get the idea. Continue until you find a difference between energy levels that equals the energy you calculated in the first step for the radiation of 4.57 x 10^14 Hz.

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