Radioactive decay question Ra to Rn

[rshalloo]

Homework Statement

Ra (mass number 226 atomic number 88) undergoes alpha decay to form radon. If the mass of a radium nucleus is 3.753152x10^-25kg, the mass of the radon nucleus is 3.686602x10^-25kg and the mass of the alpha particle is 6.646322x10^-27kg, find the energy released in the alpha decay in MeV.

If the ratio of the mass of the products of the decay is equal to the ratio of their mass numbers, find the kinetic energy of both products after the decay

charge of electron=1.6x10^-19C
speed of light in a vacuum=3x10⁸ms^-1

The Attempt at a Solution

So for the first part I got .488MeV using the mass defect and E=mc² (could well be wrong) but I am stuck on the second part i don't understand what they are talking about ratios for?

 

[kntsy]

Homework Statement

Ra (mass number 226 atomic number 88) undergoes alpha decay to form radon. If the mass of a radium nucleus is 3.753152x10^-25kg, the mass of the radon nucleus is 3.686602x10^-25kg and the mass of the alpha particle is 6.646322x10^-27kg, find the energy released in the alpha decay in MeV.
If the ratio of the mass of the products of the decay is equal to the ratio of their mass numbers, find the kinetic energy of both products after the decay
charge of electron=1.6x10^-19C
speed of light in a vacuum=3x10⁸ms^-1

$${Ra}_{88}^{226}\rightarrow{\alpha}_2^4+{Rn}_{86}^{222}$$

$$\text{energy released}=\frac{(3.753152\times{10}^{-25}-3.686601\times{10}^{-25}-6.646322\times{10}^{-27})(3\times{10}^8)^2}{1.6\times10^{-19}}=488MeV$$$$KE=\frac{mv^2}{2}=\frac{m^{2}v^2}{2m}$$

$$\text{By conservation of linear momentum}, m_{\alpha}v_{\alpha}=m_{Rn}v_{Rn}$$

$$\therefore\ \ \ \ \ \frac{KE_{\alpha}}{KE_{Rn}}=\frac{m_{Rn}}{m_\alpha}=\frac{222}{4}$$

I think all the decay energy goes to KE. Then:$$KE_\alpha=488MeV\times\frac{222}{222+4}=479MeV\ \ \ \ \ \ \ \ \ \ \ \ KE_{Rn}=9MeV$$