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Beef with my last test

  1. Dec 18, 2004 #1
    I've got some beef with how my last Physics III test was graded. I was hoping some of you might work out the following problems and post your answers to see if I had the right answers. Thanks!

    1. Potassium 40 disintegrates by emitting a positron. What is the maximum energy for the emitted positron? Given information: potassium 40 mass is 39.96400 Da, rgon 40 mass is 39.962384 Da, and the electron mass is 5.4858 x 10^-4 Da.

    EVERYBODY in the class got the same answer for this problem, and EVERYBODY was awarded 2 points out of 10, so I'm really curious to see if maybe the professor got the wrong answer.

    2. A piece of charcoal contains 250 grams of carbon. It was found during an archaeological excavation. The carbon 14 activity was measured as 250 disintegrations in 1 minute. How old is this charcoal specimen? Given information: the half-life of carbon 14 is 5730 years, and the original fraction of carbon 14 was 1.3 x 10^-12.

    I triple-checked my answer, and it also makes sense by estimation. Any help?

    3. Two protons are moving with equal speeds but in directly opposite directions. They continue to exist after a head-on collision that produces a neutral pion. If the two protons and the pion are all at rest after the collision, find the initial speed of the protons. Energy is conserved in this collision. The mass of the pion is 2.40 x 10^-24 kilograms.

  2. jcsd
  3. Dec 18, 2004 #2
    Hi rjd1, welcome to Physics Forums!
    Why don't you write up what you did and post it. Then people can critique what you did. You will find that you will recieve much more help that way. This is due to the fact people don't want to feel like they are simply doing your work for you.
  4. Dec 18, 2004 #3

    I had a suggestion to explain what I did to everybody (Thanks, Norman). Hope I don't get embarrassed. Here it is:

    1. The idea I used was

    [rest energy of potassium 40] - [rest of argon 40 and positron] = [maximum possible energy of positron]

    Plugging in the numbers, I got

    (39.964 u)(931.5 MeV/u) - (39.962384 u + 5.4858 x 10^-4 u)(931.5 MeV/u) = 37226.466 MeV - 37225.4717 MeV = 0.9943 MeV

    2. Here I'll use '14C' to denote carbon 14. I also assumed that the given fraction of 14C was a mass fraction, which I'm pretty sure is what the teacher intended. First I found the disintegration constant using the given half-life T[half] of 5730 years.

    [disintegration constant] = ln(2)/T[half] = 1.210 x 10^-4 inverse years = 2.302 x 10^-10 inverse minutes

    Then I used that disintegration constant and the given rate R of disintegration to find the current number N of 14C nuclei in the sample.

    N = R/[disintegration constant]

    N = (250 inverse minutes)/(2.302 x 10^-10 inverse minutes) = 1.08 x 10^12

    I also calculate the given "original" number N[o] of 14C nuclei in the sample.

    N[o]= (250 g C) (1.3 x 10^-12 g 14C/g C) (1 mol 14C/14 g 14C) (6.02 x 10^23 14C nuclei/mol 14C) = 1.398 x 10^13

    So then I used N, N[o], and the disintegration constant to find the age, or time t, of the sample.

    t = -ln(N/N[o])/[disintegration constant] = 2.555/(1.210 x 10^-4 inverse years) = about 21,000 years

    Scratch number 3 for now. I'll be happy if I get any responses to these two. Thanks.
  5. Dec 19, 2004 #4


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    Staff: Mentor

    This would work if the isotopic masses were the masses of the bare nuclei. But they're not. They're the masses of complete (neutral) atoms. Therefore the isotopic mass of K-40 includes the mass of 19 electrons (and their binding energy, which is probably negligible here), but the mass of Ar-40 includes the mass of only 18 electrons, and you have to compensate for that.
  6. Dec 26, 2004 #5
    You're supposing all available energy went to the positron, leaving the Ar-atom with zero kinetic energy and therefor in total rest. Can you see the conservation-of-momentum-problem with that?
    I grant you that the fraction of energy given to the Ar is very small due to its high mass relative to the positron mass, but I'm pretty sure the aim of the question was to take it into account.
    Last edited: Dec 26, 2004
  7. Dec 26, 2004 #6
    So far, so good.
    Well, you have 250g which is almost entirely made up of 12C, yet your third factor is (1 mol 14C/14 g 14C).
    #mol C = 250g/(marginally different than 12g/mol)=20.8333mol C
    #mol 14C = 20.8333mol C x (1.3 x 10^-12 mol 14C/mol C) = 2.7083 x 10^-11 mol 14C [because your teacher is bound to have meant molar fraction]
    #nuclei 14C = (2.7083 x 10^-11 mol 14C) x (6.02 x 10^23 14C nuclei/mol 14C) = 1.6304 x 10^13
    You were so close.
  8. Dec 27, 2004 #7
    Thanks, everybody.
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