Calculate the kinetic energy of the alpha particle in Problem 25 . This time, do not assume that

Calculate the kinetic energy of the alpha particle in...

Key Concepts

Mass and Energy Relationship

The masses of the decay products significantly affect the kinetic energy they acquire. Typically, the lighter particle ends up with a larger share of the kinetic energy because, for a fixed momentum, kinetic energy is inversely proportional to mass.

Recoil Effect

When a decay product is not assumed to be at rest, the recoil momentum of the heavier nucleus must be taken into account. This recoil effect influences the energy balance in the system, ensuring that both products share the total kinetic energy derived from the reaction.

Conservation of Momentum

In any two-body nuclear reaction or decay, the total momentum before and after the event must be equal. This principle is essential in determining the velocities—and thereby the kinetic energies—of the products, as the momentum lost by one particle must be gained by the other.

Kinetic Energy Distribution in Two-Body Decay

In a two-body decay, energy is partitioned between the two resulting fragments based on both their masses and the overall energy released. By applying conservation of momentum together with energy conservation, one can derive the ratio of the kinetic energies between the decaying products.

Transcript

00:01 Hello, and in this question here, we're going to be looking at the radioactive decay of radium into radon while emitting an alpha particle.

00:11 So just for convenience later on the question, we're going to need the mass of the alpha particle, the radon and the radium.

00:20 And from the appendices in the back of the book, these are the figures here.

00:24 And u is defined as 112 of the mass of a carbon 12 nuclei, and this is a value of 1 .6.

00:32 6 times 10 to the minus 27 kilograms.

00:37 So in all radiobacter decays, momentum and energy are conserved.

00:43 And we're going to assume in the beginning of the question that this radium is at rest.

00:51 Okay? so using the fact that we've always observed momentum and energy to be conserved, we can use this to construct two equations to help us find the kinetic energy of the alpha particle, which is what we're looking for in the question.

01:08 So using the fact that momentum is conserved, well, initially if the radium is at rest, the velocity is zero, so therefore the initial momentum must be equal to zero.

01:18 And the final momentum must be equal to the velocity of the radium times the mass of the radon, sorry, the final velocity must be equal to the mass of the radon times the velocity of the radon plus the mass of the alpha part.

01:34 Times the velocity of the alpha particle.

01:37 We can rearrange this to guess that the mass of the radon is equal to the mass of the alpha particle divided by the mass of the radon times the velocity of the alpha particle and we also go minus sign.

01:52 Okay? now in previous questions and question 25 in particular, we said that this velocity here was zero and our reasoning or some verification why we use said this is because the mass of the alpha particle divided by the mass of the radium is a very small number.

02:12 So we've got a very small number multiplying the velocity of the alpha particle.

02:18 So we conclude that the mass, that the velocity of the radium is approximately zero.

02:24 However, this is an approximation and we're now doing it for the case when we're assuming that the velocity of the radium is no longer zero.

02:33 So that was just a verification or some evidence to why that might be a reasonable assumption.

02:43 So we also have conservation of energy and the initial energy is equal to the mass of the radium times c squared.

02:52 And then the final energy, okay, well, that is equal to the mass of the alpha particle plus the mass of the radon times c squared.

03:01 So so far we've just included terms from the mass energy equivalence from einstein's e equals mc squared.

03:09 But we also have kinetic energy terms in the final equation...

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