AP Physics Question!!

Question

A nucleus of mass 228 u, initialy at rest, undergoes alpha decay. If the alpha particle emitted has a kinetic energy 4.7 Mev, what is the kinetic energy of the recoiling daughter nucleus? Answer in units of Mev.

Answer 1

This is a momentum conservation question funnily enough.

Since the atom was at rest before the decay we know in the original momentum is 0, and will be 0 after the decay as well.

So if the net momentum must be zero, then we know the momentum of the daughter nucleus and the helium nucleus must be the same.

we can use the equation T = P² /(2m), since we are doing alpha decay m = 4u for the helium nucleus.

So: 4.7 Mev = P²/(8u) so P = 6.1318 with the appropriate units.

This is the same momentum the daughter nucleus has after to conserve momentum

using the equation again we have KE = 6.1318²/224 = 0.167 Mev

hope this helps!

Answer 2

The best approach would be to use the law of conservation of momentum because momentum is always conserved.

Initial momentum of system is zero.

Final momentum = momentum of alpha particle + (momentum of daughter nucleus). The negative in your answer signifies a recoil or that the two masses move in opposite direction after the interaction.

Now using the KE of the alpha particle determine its velocity. KE = 1/2mv²

Now you have the mass of the α-particle, the velocity of the α-particle and the mass of the daughter nucleus. (228-4u). Use these values in the expression below

0 = P₁ + P₂, or P₁ = - P₂ I will let you do the calculations.

It is possible to use the conservation of energy instead of momentum but since we are not told that the event is elastic (and elastic interactions are rare in reality), we should not make that assumption.

I hope this helps.