Nuclear Physics: alpha decay and kinetic energy...?

Question

A nucleus of mass 228 u, initially at rest, undergoes alpha decay. If the alpha particle emitted has a kinetic energy of 4.00 Mev, what if the kinetic energy of the recoling daughter nucleus?

answer in book: 71.4 keV

how do you solve this kind of problem? what equations should i use and what steps to follow? HELP! thanks

Answer 1

this is a basic conservation of moment question i think.

a alpha particle has mass of roughly 4u which is 6.644656×10-27Kg

1 u = 1.661×10−27 kg

convert 4.00Mev into joules

4000000eV =4000000 x 1.602×10-19 J

now use E(k) = 1/2 mv^2 to find v

so u know the alpha particle has momentum of mv  (its mass x its velocity)

work out the mass of the nuclei (228u-4u)

convert 224u into Kg

the nucleis momentum = the alpha particles momentum.

so its mass x velocity = the alpha particles mass x velocity

so its velocity (ms^-1)  = (mass of alpha particle x its velocity)/the mass off the nuclei) to find the velocity of the nuclei

now using E(k) = 1/2 mv^2 sub this value of v into it, to find out how much energy it the nucleus has

Answer 2

Yes, darealbud is right, it does use conservation of momentum - but I think we can simplify it down a bit:

Since momentum is conserved, the momentum of the alpha particle is the same as the momentum of the daughter nucleus, just in the opposite direction.

1. Let 0 represent the alpha particle, and 1 represent the daughter nucleus:

m0 * v0 = -m1 * v1

2. Kinetic energy E = (1/2)*(m)*(v^2), so:

v = sqrt(2*E/m)

3. Substituting back into the momentum equation:

m0 * sqrt(2*E0/m0) = -m1 * sqrt(2*E1/m1)

4. Square both sides:

m0^2 * (2*E0/m0) = m1^2 * (2*E1/m1)

5. Once you cancel out variables, you can finally get:

(m0 / m1) * E0 = E1

Since you know m0, m1 and E0, you can get E1.

Hope this helps!  ^_^