Decay of Radium Finding Kinetic Energy of products

[rwooduk]

Homework Statement

Homework Equations

Please see below.

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?

 

[mfb]

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.

 

[rwooduk]

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.

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.

 

[mfb]

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

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

 

[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.

 

[rwooduk]

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