Concept explainers
(a)
To Calculate: Macroscopic capture cross section.
(a)
Answer to Problem 5.40P
Macroscopic capture cross section
Explanation of Solution
Given:
Fe | Cr | Ni | ||||||
A | Abun(%) | A | Abun(%) | A | Abun(%) | |||
54 | 5.84 | 2.3 | 50 | 4.35 | 15.9 | 58 | 68.08 | 4.6 |
56 | 91.75 | 2.6 | 52 | 83.79 | 0.75 | 60 | 26.22 | 2.8 |
57 | 2.12 | 2.4 | 53 | 9.50 | 18.1 | 61 | 1.14 | 2.5 |
58 | 0.28 | 1.1 | 54 | 2.37 | 0.36 | 62 | 3.635 | 14.4 |
64 | 0.926 | 1.5 |
Weight percent:
Fe = 71%
Cr = 19%
Ni = 10%
Formula used:
Calculation:
Using the above equation calculation was done:
1 | 2 | 3 | 4 | 5 | 6 | 7 |
Fe | 7.874 | 54 | 0.0584 | 2.3 | 0.0117946 | |
Fe | 7.874 | 56 | 0.9175 | 2.6 | 0.20198892 | |
Fe | 7.874 | 57 | 0.0212 | 2.4 | 0.00423261 | |
Fe | 7.874 | 58 | 0.0028 | 1.1 | 0.0002518 | |
total | 0.21826793 | |||||
Fe total | 0.15497023 | |||||
Cr | 7.19 | 50 | 0.0435 | 15.9 | 0.05989437 | |
Cr | 7.19 | 52 | 0.8379 | 0.75 | 0.05232627 | |
Cr | 7.19 | 53 | 0.095 | 18.1 | 0.140474 | |
Cr | 7.19 | 54 | 0.0237 | 0.36 | 0.00068411 | |
total | 0.25337875 | |||||
Cr total | 0.04814196 | |||||
Ni | 8.908 | 58 | 0.6808 | 4.6 | 0.28964787 | |
Ni | 8.908 | 60 | 0.2622 | 2.8 | 0.06563877 | |
Ni | 8.908 | 61 | 0.0114 | 2.5 | 0.00250632 | |
Ni | 8.908 | 62 | 0.03635 | 14.4 | 0.04528936 | |
Ni | 8.908 | 64 | 0.00926 | 1.5 | 0.00116424 | |
total | 0.40424656 | |||||
Ni total | 0.04042466 | |||||
TOTAL | 0.24353685 |
Conclusion:
Macroscopic capture cross section
(b)
The number of captured neutrons per second
(b)
Answer to Problem 5.40P
The number of captured neutrons per second
Explanation of Solution
Given:
Formula used:
Calculation:
From the above equation,
Therefore,
1 cm diameter correspond to an area of
The number of captured neutrons can be calculated as,
Conclusion:
The number of captured neutrons per second
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Chapter 5 Solutions
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