Solution:
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(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
.
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