using special relativity m = m0/(1-v^2/c^2) on an electron orbiting with a speed of 2.01 * 10^8 m/s in an atom,estimate the percentage corrections to the n=1 energy in a uranium (Z=92) atom.(adsbygoogle = window.adsbygoogle || []).push({});

so to find the energy I first need the reduced mass, u, for En = u*(9e9)^2*e^4*Z^2/2(h bar)^2*n^2.

How do I find the reduced mass? I know it is u = Me * M(nucleus)/(Me + Mnucleus)

but how can I find the mass of the nucleus. The answer my professor gave was there was a correction of 35%, but I cannot compare the energies until I have the reduced mass that will be in the first energy, then I change the reduced mass using the special relativity formula above, and then find the n= 1 energy again and find the percent difference.

I would really appreciate any help.

Thanks.

Stephen

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# Homework Help: Energy of an electron in a uranium atom

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