Kinetic energy of alpha particle in alpha decay

[Kara386]

Homework Statement

Consider the nuclear decay ²¹⁴₈₆Rn → ²¹⁰₈₄Po^2- + α.
Calculate the Q for this decay, and give the value of the kinetic energy of the alpha particle in the rest frame of the Rn nuclide. The rest mass of the Rn nuclide is 213.9954u, of Po is 209.9829u, of α is 4.00015u and of an electron is 0.0005u. Neglect relativity!

Homework Equations

$Q = (m_P - m_D-m_{\alpha})c^2$
$m_P$ is mass of parent, $m_D$ is mass of daughter, $m_\alpha$ is mass of the alpha particle (rest masses, I think)

The Attempt at a Solution

The equation above gives a Q value of 10.2465MeV, assuming that because masses are in units of u then $c^2=931.5MeV/c^2$. Or that's what I thought, until I noticed the $2-$ superscript for Po. Should I add two lots of the electron mass onto my answer? Not that it makes a massive difference, but since we're given the electron mass presumably I have to use it.

For the second part, label kinetic energy of $\alpha$ particle $T_{\alpha}$, of daughter nuclide $T_D$. I'm going to say that since we're in the rest frame of the Rn nuclide, that means it was initiallty at rest? So by conservation of momentum,

$p_{\alpha}+p_D = 0$ and $p_{\alpha} = \sqrt{2m_{\alpha}T_{\alpha}}$ ;
$p_{D} = -\sqrt{2m_{D}T_{D}}$. So since $p_D = -p_{\alpha}$:
$T_{\alpha} = \frac{m_D}{m_{\alpha}}T_D$
That doesn't really get me anywhere, I don't think. Is this the wrong approach? Where could I go from here? Thanks for any help!

 

[mfb]

You'll have to include the electron mass, and it gives about 10% difference.

That doesn't really get me anywhere, I don't think.

Why not? What can you say about $T_\alpha + T_D$?

 

[Kara386]

You'll have to include the electron mass, and it gives about 10% difference.
Why not? What can you say about $T_\alpha + T_D$?

10%? Oops, that is pretty big. I'm not 100% sure what Q actually is.Some sort of energy; is it equal to $T_{\alpha}+T_D$?

 

[mfb]

Well, your Q-value was 10 MeV, and the mass of two electrons is about 1 MeV.

Q: Right. Energy is conserved, so the total energy of the final state (masses+kinetic energies) has to match the total energy of the initial state (radon mass).