# Decay of Radium Finding Kinetic Energy of products

1. May 28, 2015

### rwooduk

1. The problem statement, all variables and given/known data

2. Relevant equations

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

$$COM \\ \\ E_{6}= E_{2}+E_{\alpha} \\ \\ Minus \ E_{2} \ and \ square \ both\ sides\\ \\ E_{6}^{2}+E_{2}^{2}-2E_{2}E_{6}=E_{\alpha}^{2}\\ \\ E_{6}^{2}= m_{6}^{2}c^{4}\\ \\ E_{2}^{2}= \rho_{\alpha}^{2}c^{2} + m_{2}^{2}c^{4}\\ \\ E_{\alpha}^{2}= \rho_{\alpha}^{2}c^{2} + m_{\alpha}^{2}c^{4}\\ \\ Rearranging \ I \ get: \\ \\ E_{2}=\frac{(m_{6}^{2}+m_{2}^{2}-m_{\alpha}^{2})c^{4}}{2E_{6}}$$

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. May 28, 2015

### 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.

3. May 28, 2015

### rwooduk

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.

4. May 28, 2015

### Staff: Mentor

That part cannot be right, the units do not match. I guess the energy should be squared.

5. May 29, 2015

### theodoros.mihos

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.

6. May 30, 2015

### rwooduk

Yes, thank you both, you are right I have it now!