A thermal neutron beam of intensity 1 to the power 11, neutrons/sec is directed for 2 days into a cavity in a block of cobalt initially composed entirely of the stable isotope Cobalt 59. The neutrons are all captured, producing cobalt 60 nuclei which are radioactive with a half-life of 5.3 years.
What is the activity of the cobalt 60
(y-ray) source produced at the end of the two days?
To calculate the activity of the cobalt 60 (γ-ray) source produced at the end of the two days, we need to use the concept of radioactive decay and the decay constant.
First, let's calculate the number of cobalt 60 nuclei produced. Since all the neutrons are captured and cobalt 60 is produced, the number of cobalt 60 nuclei is equal to the number of neutrons in the beam.
Given:
Intensity of the neutron beam = 1 x 10^11 neutrons/sec
Duration of the beam directed into the cavity = 2 days = 2 x 24 x 60 x 60 seconds
Number of neutrons in the beam:
Number of neutrons in the beam = Intensity x Duration
Number of neutrons in the beam = 1 x 10^11 neutrons/sec x 2 x 24 x 60 x 60 seconds
Now, we need to determine the number of cobalt 60 nuclei produced. Since cobalt 60 has a half-life of 5.3 years, we can convert the given duration of 2 days into years and use the radioactive decay formula.
Number of cobalt 60 nuclei produced:
Number of cobalt 60 nuclei produced = Number of neutrons in the beam
Next, we calculate the activity of the cobalt 60 γ-ray source.
Activity is given by the decay constant (λ) times the number of cobalt 60 nuclei.
Decay constant (λ) is related to the half-life (T1/2) by the equation:
λ = ln(2) / T1/2
The decay constant (λ) for cobalt 60 can be calculated using the half-life provided.
Finally, we can calculate the activity:
Activity = λ x Number of cobalt 60 nuclei produced
By following these steps, you can calculate the activity of the cobalt 60 (γ-ray) source produced at the end of the two days.