Chapter 5, Problem 5.40P

(a)

To determine

To Calculate: Macroscopic capture cross section.

Answer to Problem 5.40P

Macroscopic capture cross section $=0.243c{m}^{-1}$

Explanation of Solution

Given:

Fe			Cr			Ni
A	Abun(%)	${\sigma }_C$ (b)	A	Abun(%)	${\sigma }_C$ (b)	A	Abun(%)	${\sigma }_C$ (b)
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:

  $\begin{array}{l}{\displaystyle \sum =\frac{\rho \times 1mole\times \sigma \times (percentabundence)}{gramatomicwgt}}\\ {\displaystyle \sum =\frac{\rho \times 1mole\times \sigma \times b\times 1\times {10}^{-24}\times (fractionalabundence)}{gramatomicwgt\times b}}\end{array}$

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 $=0.243c{m}^{-1}$

(b)

To determine

The number of captured neutrons per second

Answer to Problem 5.40P

The number of captured neutrons per second $=1.96\times {10}^{10}n/\sec$

Explanation of Solution

Given:

  $\begin{array}{l}{I}_0=5\times {10}^{11}neutrons/c{m}^2\cdot s\\ t=0.2cm\\ {\displaystyle \sum =0.243c{m}^{-1}}\end{array}$

Formula used:

  $\frac{I}{I_0}=e^{-\sum t}$

Calculation:

From the above equation,

  $\begin{array}{l}\frac{I}{I_0}=e^{-\sum t}\\ =e^{-0.243\times 0.2}\\ =0.95\end{array}$

Therefore, $1-0.95=0.05$ are captured

1 cm diameter correspond to an area of $=0.7854c{m}^2$

The number of captured neutrons can be calculated as,

  $\begin{array}{l}=0.05\times 5\times {10}^{11}\times 0.7854\\ =1.96\times {10}^{10}n/\mathrm{s}\end{array}$

Conclusion:

The number of captured neutrons per second $=1.96\times {10}^{10}n/\mathrm{s}$