Rydberg equation

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Rydberg equation

Postby Chem_Mod » Sun Dec 04, 2011 2:10 am

I used the Rydberg equation for n=4 and converted frequency to energy with the E=hv formula. I have 9.27x10^-20= -3.44x10^-34 - E(initial). However when i add 9.27x10^-20 to 3.44x10^-34 i get 9.27x10^-20. I get the same number, is that correct?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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Re: Rydberg equation

Postby Chem_Mod » Sun Dec 04, 2011 2:10 am

When I solved the equations, I got different numbers. Determining the energy of the photon requires the formula E=hv: E(photon) = -(6.626 x 10^-34 Js)(1.14 x 10^14 Hz). You have to realize that since the photon is being emitted, it is a negative energy value, so you must make your answer negative. If the photon was being absorbed, its energy would be positive. Then use the Rydberg equation (for an H atom) to find the energy of the n=4 state: E(n=4) = -(hR)/n^2 E(n=4) = -[(6.626 x 10^-34 Js)(3.29 x 10^15 Hz)]/(4^2) We already talked about the equation E(photon) = E(final) - E(initial) and E(n=4) is the final energy. You solve for E(initial) and then use the Rydberg equation to find n for that energy: E(n=?) = -(hR)/n^2 n^2 = -(hR)/E(n=?) Take the square root and you should get close to an integer. Your energy values for both levels and the photon should be on the order of 10^-18 to 10^-20 J. If you are getting answers on the order of 10^-34 J, you need to redo the math.


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