1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Decay of Radium Finding Kinetic Energy of products

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Please see below.

    3. The attempt at a solution
    I'm pretty sure my method is correct but I'm getting small answers:

    COM \\ \\

    E_{6}= E_{2}+E_{\alpha}
    Minus \ E_{2} \ and \ square \ both\ sides\\

    E_{6}^{2}+E_{2}^{2}-2E_{2}E_{6}=E_{\alpha}^{2}\\ \\

    E_{6}^{2}= m_{6}^{2}c^{4}\\

    E_{2}^{2}= \rho_{\alpha}^{2}c^{2} + m_{2}^{2}c^{4}\\ \\
    E_{\alpha}^{2}= \rho_{\alpha}^{2}c^{2} + m_{\alpha}^{2}c^{4}\\ \\

    Rearranging \ I \ get: \\ \\

    Now this is where I get strange results (probably my units), if I sub in for $$E_{6}$$ and insert numerical values I get:

    $$E_{2}=\frac{(8.71\cdot 10^{10})(\frac{MeV}{c^{2}})^{2}c^{4}}{2\cdot (210541.379)(\frac{MeV}{c^{2}})c^{2}}= \frac{(8.71\cdot 10^{10})(\frac{MeV}{c^{2}})c^{2}}{2\cdot (210541.379)}$$

    Which is 206808MeV. Then if I divide by c^2 I get ~7x10^-4 MeV/c^2 which is really small?

  2. jcsd
  3. May 28, 2015 #2


    User Avatar
    2017 Award

    Staff: Mentor

    What do you get for the kinetic energy of the alpha particle?
    Does the energy balance work out with those numbers?

    If yes, what are the velocities of the particles - does that conserve momentum?

    If not, something went wrong. Probably in the "rearranging" part that you did not show.
  4. May 28, 2015 #3
    I get $$E_{\alpha}= \sqrt{E_{6}^{2}+E_{2}^{2}-2E_{6}E_{2}}$$

    Now I'm really confused (it's getting late in the day!) because:

    $$E_{6}= m_{6}^{2}c^{4}= (210541.379)^{2}(\frac{MeV}{c^{2}})^{2}c^{2}=147(\frac{MeV}{c})^{2}$$

    Why do I have (MeV)^2?

    Think I'll start this one fresh in the morning, I think I'm getting the units all mixed up. Not sure why I divided by c^2 at the end of the OP.
  5. May 28, 2015 #4


    User Avatar
    2017 Award

    Staff: Mentor

    That part cannot be right, the units do not match. I guess the energy should be squared.
  6. May 29, 2015 #5
    Before decay you have ## \mathbf{p}_0 = 0 ## and ## E_{Ra} ## (on cm frame) . After decay, you have ## \mathbf{p}_a + \mathbf{p}_{Rn} ## and ## E_{a} + E_{Rn} ##. All are vector components so, as @mfb say, may be squared.
  7. May 30, 2015 #6
    Yes, thank you both, you are right I have it now!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Decay of Radium Finding Kinetic Energy of products