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Energy of an electron in a uranium atom

  1. Jan 18, 2012 #1
    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.

    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
     
  2. jcsd
  3. Jan 18, 2012 #2
    Would the reduced mass be = (Me * (92*Mproton))/(Me +92*Mproton)???

    Please, a quick reply would be greatly appreciated.

    Thanks.
    Stephen
     
  4. Jan 18, 2012 #3
    Just a quick aside as a part of the same problem for a hydrogen at n=1 state with a reduced mass of 9.10444*10^(-31) kg with special relativistic effects makes u = 9.10468*10^(-31) kg
    to make E =u*(8.988*10^9)^2*(1.6022*10^(-19))^4/ (2(1.0546*10^(-34))^2)

    which makes the percent correction for hydrogen .0027%.

    is this right???
     
  5. Jan 18, 2012 #4

    Dick

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    Actually you would use mass of the uranium nucleus in that formula. That would be about 238*Mproton because there are a lot of neutrons in there too. But you needn't bother. The reduced mass correction is insignificant compared with the relativistic correction in uranium. Just compute that one.
     
  6. Jan 18, 2012 #5
    How would I find the relativistic mass without knowing the reduced mass?? What formula should I use to find the reduced mass?

    Thanks
    Stephen
     
  7. Jan 18, 2012 #6

    Dick

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    You have the right formula for the reduced mass. You should use the mass of the U nucleus instead of Mproton. But you are going to find the reduced mass is really really really close to being the same as Me because the nucleus is so heavy. The important mass correction is the one you had in your problem statement m = m0/sqrt(1-v^2/c^2). I just noticed you left the square root out of that formula. Just find that correction.
     
  8. Jan 18, 2012 #7
    so use m0=Me is this what you are saying??
     
  9. Jan 18, 2012 #8

    Dick

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    That's exactly what I'm saying.
     
  10. Jan 18, 2012 #9
    ok...so I emailed my professor and he said to use the same numbers as in the following problem:

    The problem was to find the speed of an electron in a uranium atom in n=1.
    so I started with r = (hbar)^2/ u*(9e9)(92)e^2=5.7549*10^-13 m with u = 9.10444*10^(-34) kg which is the reduced mass of a hydrogen atom then I went to F=ke^2/r^2 = mv^2/r and found v^2 = 2.10 * 10^7 m/s but my professor says that it is 2.01*10^8 m/s.

    So I do not have the numbers right in this problem and so my numbers for the problem of the thread is wrong. Where am I messing up??

    Please, Please help. My professor just said that I should think about it and the fact that hydrogen is Z=1 and uranium is Z=92.

    Thanks.
    Stephen
     
  11. Jan 18, 2012 #10

    Dick

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    I'm not following all of that, but when you do F=ke^2/r^2 = mv^2/r it looks like you are balancing electric force and centripetal acceleration. But why ke^2/r^2? Isn't the the charge of the nucleus 92e, not e?
     
  12. Jan 18, 2012 #11
    ahhh...you are right. so would the force =k92e^2/r^2=mv^2/r?? So I do not need to worry about the reduced mass, just use the mass of electron,9.1094*10^(-31) kg, or the reduced mass of an electron proton pair, 9.10444*10^(-31) kg??

    Thanks.
    Stephen
     
  13. Jan 18, 2012 #12

    Dick

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    Yes, use Me. Since you don't seem to believe me find the reduced mass in uranium. Tell me how close it is to Me. There is absolutely no reason to use the reduced mass of an electron proton pair. That has nothing to do with this problem at all.
     
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