Radon decays as ^222 Rn → ^218 Po+α. The mass of the radon- 222 nucleus is 221.97039 u, the mas

Radon decays as ^222 Rn → ^218 Po+α. The mass of the radon-...

Key Concepts

Kinetic Energy Partitioning in Two-Body Decay

In a two-body decay, the kinetic energy generated (Q-value) is divided between the two particles inversely proportional to their masses because they must share equal and opposite momentum. The particle with the smaller mass generally receives a larger share of the kinetic energy, which directly influences its speed relative to the heavier particle.

Mass-Energy Equivalence

Mass-energy equivalence, encapsulated by Einstein’s relation E=mc², underpins the conversion of the mass defect into energy during a nuclear decay. The conversion factor (1 u = 931.5 MeV/c²) allows one to compute the Q-value, effectively linking a small loss in mass to a measurable release of kinetic energy in the decay process.

Non-relativistic Kinematics

For many nuclear decay processes, the kinetic energy amounts are small in comparison to the rest energy of the particles, permitting the use of non-relativistic formulas like K = ½ mv² to determine the speeds. This approximation simplifies calculations while still providing accurate results given the relatively low velocities involved in such decays.

Conservation of Momentum

In two?body nuclear decays that start from a stationary parent nucleus, momentum must be conserved. This means that the momenta of the resulting particles are equal in magnitude and opposite in direction, ensuring that the net momentum remains zero. This principle is crucial when determining the speeds of the decay products by linking their masses to their velocities.

Conservation of Energy

During a nuclear decay, the total energy is conserved. The difference in mass between the initial nucleus and the decay products is converted into kinetic energy of the emitted particles. The Q-value, obtained from the mass defect multiplied by the conversion factor, represents the total kinetic energy gained by the products, and this energy is partitioned according to their masses.

Transcript

00:01 In this problem, we are looking at the decay of radon from rest into polonium and an alpha particle, which is a helium nucleus.

00:08 This actually is a paired question with problem 85, the previous question.

00:12 This is exactly the same problem up until a certain point.

00:16 There's two distinct differences, one minor, one major.

00:19 First one is we have different units for the mass.

00:23 Second one is that it does not stop at just wanting the kinetic energy of the decay products, but it wants the speed.

00:29 So that's really the new part of the question.

00:31 This really is a question 85 second time with those add -ins.

00:37 What do we do here as we did there? we have our energy conservation equation, including rest energy and kinetic energy and moving decay products.

00:46 We have our momentum conservation equation.

00:49 And then from that, we can, momentum conservation, we can write expression that involves the relationship between the momentum and the kinetic energy.

01:00 So we have two equations, two unknowns.

01:03 And what do you normally do? you solve one for one of the unknowns and place it into the other.

01:07 And that's what we do here.

01:09 And when all said and done, the end result, for in this case, the kinetic energy of the polonium is this expression.

01:18 And i'll come back to that in a second.

01:20 I just want to make one comment, reminder.

01:24 This quantity here that i defined as m.

01:26 This is the amount of rest mass that is lost in the decay that shows itself as the total kinetic energy of the decay products.

01:38 K -alpha plus k -p -polonium is equal to this quantity.

01:44 M -c -squared.

01:45 That tells you whatever you lost in rest mass went into kinetic energy.

01:52 So now let's get back.

01:54 So like all the all the algebraic steps are laid out here.

01:57 There's nothing, nothing that is left out.

01:59 So that's easy enough to follow this again at this stage up until this point it really is no different than question 85 exactly the same analysis so now let's look at units and calculate mc squared 221 .9703 you minus 217 .9628 9u minus 4 .00151.

02:38 U times c squared.

02:41 This works out to be 5 .99 times 10 to the minus 3, uc squared.

02:49 Uc squared is in combination energy, units of energy when you put them together.

02:56 Do i have to convert that? the problem gives us conversion between u and mev.

03:02 Mvv per c squared to be precise can uh do i need to convert this to m eve i could certainly if i want to but i don't have to it's units this together is energy units of energy just like you have this weird unit for for power kilowatt hour that's not power that is that is actually energy kilowatt is the power and then you have the hour that's the time so that gives you energy it's a weird it's a weird unit, but it's something you've dealt with.

03:35 So that is that formula, to that number.

03:40 So we can now immediately go and use their formula to get the kinetic energy of the polonium.

03:47 5 .99 times 10 to the minus 3 u .c squared...

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