Problem #0148 Quantum Physics Problem #0150 Chapters Chapters

850.

Problem 54.67 (RHK)

Two radioactive materials that are unstable to alpha decay, and , and one that is unstable to beta decay, , are sufficiently abundant in granite to contribute significantly to the heating of the Earth through the decay energy produced. The alpha-unstable isotopes give rise to decay chains that stop at stable lead isotopes. has a single beta decay. The decay information is as follows:

is the total energy released in the decay of one parent nucleus to the final stable endpoint and is the abundance of the isotope in kilograms per kilogram of granite; ppm means parts per million. We have to show (a) that these materials give rise to a total heat production of 988 pW for each kilogram of granite. (b) Assuming that there is of granite in a 20-km thick, spherical shell around the Earth, we have to estimate the power this will produce over the whole Earth. We will compare this with the total solar power intercepted by the Earth, .

Solution:             Click For PDF Version

(a)

:

As the abundance of in 1 kg of granite is , the number of nuclides in 1 kg of granite will be

The half-life of for radioactive decay to stable endpoint is . The disintegration constant will be

The decay rate of nuclides in 1 kg of granite will be

For this decay is 51.7 MeV. Therefore, energy released per second due to radioactive decay of nuclides contained in 1 kg of granite will be

:

We consider next the energy production in 1 kilogram of granite because of radioactive decay of nuclides contained in it.

As the abundance of in 1 kg of granite is , the number of nuclides in 1 kg of granite will be

The half-life of for radioactive decay to stable endpoint is . Its disintegration constant will be

The decay rate of nuclides in 1 kg of granite will be

For this decay is 42.7 MeV. Therefore, energy released per second due to radioactive decay of nuclides contained in 1 kg of granite will be

:

We consider next the energy production in 1 kilogram of granite because of radioactive decay of nuclides contained in it.

As the abundance of nuclides in 1 kg of granite is , the number of nuclides in 1 kg of granite will be

The half-life of for beta decay is . The disintegration constant will be

The decay rate of nuclides in 1 kg of granite will be

For this decay is 1.32 MeV. Therefore, energy released per second due to radioactive decay of nuclides contained in 1 kg of granite will be

We thus find that the total energy released per second in 1 kilogram of granite due to radioactive disintegrations will be

(b)

Assuming that there is of granite in a 20-km thick, spherical shell around the Earth, we find that the power produced over the whole Earth because of radioactive decays will be

The total solar power intercepted by the Earth is, which is .