Kinetic Energy for particles (alpha decay)

[shepherd882]

Homework Statement

Many heavy nuclei undergo spontaneous "alpha decay," in which the original nucleus emits an alpha particle (a helium nucleus containing two protons and two neutrons), leaving behind a "daughter" nucleus that has two fewer protons and two fewer neutrons than the original nucleus. Consider a radium-220 nucleus that is at rest before it decays to radon-216 by alpha decay.

The mass of the radium-220 nucleus is 219.96274 u (unified atomic mass units) where
1 u = 1.6603 ✕ 10−27 kg
(approximately the mass of one nucleon).

The mass of a radon-216 nucleus is 215.95308 u, and the mass of an alpha particle is 4.00151 u. Radium has 88 protons, radon has 86 protons, and an alpha particle has 2 protons. (Use 2.9979 ✕ 108 m/s for the speed of light.)

a) calculate the final kinetic energy of the alpha particle. for the moment, assume that its speed is small compared to the speed of light
b) calculate the final kinetic energy of the radon-216 nucleus.

Homework Equations

Einitial = Efinal + K
K = 1/2mv^2
Pinitial = Pfinal

The Attempt at a Solution

m(radium) - m(radon) - m(alpha) = K total
(219.96274)(1.6603*10^-27) - (215.95308)(1.6603*10^-27) - (4.00151)(1.6603*10^-27) = Ktotal
Ktotal = 1.3531*10^-29 J

P(radium) = P(alpha) + P(radon)
0 = m(radon)*v(radon) + m(alpha)*v(alpha)
-m(alpha)*v(alpha) = m(radon)*v(radon)
v(alpha) = [m(radon)*v(radon)]/-m(alpha) = -53.97v(radon)

Ktotal = Kalpha + Kradon
1.3531*10^-29 = 1/2*m(alpha)(-53.97v(radon))^2 + 1/2*m(radon)*v(radon)^2
1.3531*10^-29 = 9.6759*10^-24*v(radon)^2 + 1.79275*10^-25*v(radon)^2
1.3531*10^-29 = 9.8552*10^-24*v(radon)^2
v(radon)^2 = 1.373*10^-6
v(radon) = 1.17*10^-3 m/s

Kalpha = 1/2*m(alpha)*(-53.97(1.17*10^-3))^2 = 1.32*10^-29 J
Kradon = 1/2*m(radon)*v(radon)^2 = 2.46*10^-31 J

(the answer's supposed to be: Kalpha = 1.194*10^-12 J & Kradon = 2.2*10^-14 J)

thanks in advance!

 

[TSny]

m(radium) - m(radon) - m(alpha) = K total
(219.96274)(1.6603*10^-27) - (215.95308)(1.6603*10^-27) - (4.00151)(1.6603*10^-27) = Ktotal
Ktotal = 1.3531*10^-29 J

Check the units on the left side of the middle equation. Do they give you Joules?

 

[neilparker62]

Interestingly you can calculate $ΔE = Δmc^2$ and then apportion according to the following ratios:

$$KE_{alpha} = ΔE\times\frac{m_{radon}}{m_{radon}+m_{alpha}}$$
and
$$KE_{radon} = ΔE\times\frac{m_{alpha}}{m_{radon}+m_{alpha}}$$