Radium decay In this example we will apply the concept of mass defect to a radioactive decay. For the decay to be possible, the mass of the original nucleus must be greater than the sum of the masses of the decay products. For example, suppose you are given these atomic masses:

Question

Radium decay

In this example we will apply the concept of mass defect to a radioactive decay. For the decay to be possible, the mass of the original nucleus must be greater than the sum of the masses of the decay products. For example, suppose you are given these atomic masses:

[latex]{}_{88}^{226}Ra : 226.025410 u,[/latex]

[latex]{}_{86}^{222}Rn : 222.017578 u,[/latex]

[latex] {}_{2}^4He : 4.002603 u.[/latex]

Let’s show that a emission is energetically possible for radium, and then determine the kinetic energy of the emitted a particle.

Accepted Answer

SET UP First note that the masses given are those of the neutral atoms, including 88, 86, and 2 electrons, respectively, so we have the same numbers of electrons in the initial and final states. This is essential.
Next, we have to compare the mass of [latex]{}^{226}Ra[/latex] with the sum of the masses of [latex]{}^{222}Rn[/latex] and 4 He. If the former is greater, then a decay is possible.

SOLVE The difference between the mass of [latex]^{226}Ra[/latex] and the sum of the masses of [latex]^{222}Rn[/latex] and [latex]^4He[/latex] is

226.025410 u - (222.017578 u + 4.002603 u) = 0.005229 u.

The fact that this quantity is positive shows that the decay is energetically possible. The energy equivalent of 0.005229 u is

E = (0.005229 u)(931.5 MeV/u) = 4.871 MeV.

REFLECT The total mass of the system has decreased by 0.005229 u, and the corresponding increase in kinetic energy is 4.871 MeV. Thus, we expect the decay products to emerge with a total kinetic energy of 4.871 MeV. Momentum is also conserved: If the parent nucleus is initially at rest, the daughter nucleus and the α particle have momenta with equal magnitude and opposite direction. The kinetic energy is K = p²/2m, so the kinetic energy divides inversely as the masses of the two particles. The α gets 222/(222 + 4) of the total, or 4.78 MeV, equal to the observed α energy.

Practice Problem: The nuclide [latex]^{60}Co,[/latex] an odd–odd unstable nucleus, is used in medical applications of radiation. Show that it is unstable against b decay, and find the total energy of the decay products. The atomic masses are

[latex]_{ 27}^{60}Co : 59.933817 u,[/latex]

[latex]_{28}^{60}Ni : 59.930786 u.[/latex]

Answer: 2.82 MeV.

Question 30.3: Radium decay In this example we will apply the concept of ma...

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