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Sagot :
To solve this problem, let's start by breaking it down into two parts: calculating the mean free path of thermal neutrons in cadmium and then estimating the thickness of cadmium needed to remove 99% of the neutrons from a beam.
### Part 1: Calculating the Mean Free Path
1. Given Data:
- Density of cadmium, [tex]\( \rho = 8.6 \times 10^3 \ \text{kg/m}^3 \)[/tex]
- Neutron cross-section, [tex]\( \sigma = 2.4 \times 10^4 \ \text{barns} \)[/tex] (Note: 1 barn = [tex]\( 10^{-28} \ \text{m}^2 \)[/tex])
- Avogadro's number, [tex]\( N_A = 6.022 \times 10^{23} \ \text{atoms/mol} \)[/tex]
- Molar mass of cadmium, [tex]\( M = 112.41 \ \text{g/mol} \)[/tex]
2. Convert the cross-section to square meters:
[tex]\[ \sigma = 2.4 \times 10^4 \ \text{barns} = 2.4 \times 10^4 \times 10^{-28} \ \text{m}^2 = 2.4 \times 10^{-24} \ \text{m}^2 \][/tex]
3. Convert the molar mass to kilograms per mole:
[tex]\[ M_{\text{kg/mol}} = 112.41 \ \text{g/mol} \times 10^{-3} \ \text{kg/g} = 0.11241 \ \text{kg/mol} \][/tex]
4. Calculate the number density of cadmium atoms ([tex]\(N\)[/tex]):
[tex]\[ N = \left( \frac{\rho}{M_{\text{kg/mol}}} \right) N_A = \left( \frac{8.6 \times 10^3 \ \text{kg/m}^3}{0.11241 \ \text{kg/mol}} \right) (6.022 \times 10^{23} \ \text{atoms/mol}) \][/tex]
[tex]\[ N \approx 4.612 \times 10^{28} \ \text{atoms/m}^3 \][/tex]
5. Calculate the mean free path ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{1}{N \sigma} = \frac{1}{(4.612 \times 10^{28} \ \text{atoms/m}^3)(2.4 \times 10^{-24} \ \text{m}^2)} \][/tex]
[tex]\[ \lambda \approx 9.043 \times 10^{-6} \ \text{m} \][/tex]
### Part 2: Estimating the Thickness to Remove 99% of Neutrons
The attenuation of neutrons in a material can be described by the exponential attenuation law:
[tex]\[ I(x) = I_0 e^{-x / \lambda} \][/tex]
where [tex]\(I(x)\)[/tex] is the intensity of neutrons after traveling through the material, [tex]\(I_0\)[/tex] is the initial intensity, [tex]\(x\)[/tex] is the thickness, and [tex]\(\lambda\)[/tex] is the mean free path.
To remove 99% of the neutrons:
[tex]\[ \frac{I(x)}{I_0} = 1 - 0.99 = 0.01 \][/tex]
Taking the natural logarithm on both sides, we get:
[tex]\[ \ln\left(\frac{I(x)}{I_0}\right) = \ln(0.01) \][/tex]
[tex]\[ -\frac{x}{\lambda} = \ln(0.01) \][/tex]
[tex]\[ x = -\lambda \ln(0.01) \][/tex]
Since [tex]\(\ln(0.01) = -4.605\)[/tex]:
[tex]\[ x = \lambda \times 4.605 \][/tex]
Substituting the value of [tex]\(\lambda\)[/tex]:
[tex]\[ x = 9.043 \times 10^{-6} \ \text{m} \times 4.605 \][/tex]
[tex]\[ x \approx 4.165 \times 10^{-5} \ \text{m} \][/tex]
### Final Answer:
- The mean free path of thermal neutrons in cadmium is approximately [tex]\(9.044 \times 10^{-6} \ \text{m}\)[/tex].
- The thickness of cadmium required to remove 99% of the neutrons from the beam is approximately [tex]\(4.165 \times 10^{-5} \ \text{m}\)[/tex] or 41.65 micrometers.
### Part 1: Calculating the Mean Free Path
1. Given Data:
- Density of cadmium, [tex]\( \rho = 8.6 \times 10^3 \ \text{kg/m}^3 \)[/tex]
- Neutron cross-section, [tex]\( \sigma = 2.4 \times 10^4 \ \text{barns} \)[/tex] (Note: 1 barn = [tex]\( 10^{-28} \ \text{m}^2 \)[/tex])
- Avogadro's number, [tex]\( N_A = 6.022 \times 10^{23} \ \text{atoms/mol} \)[/tex]
- Molar mass of cadmium, [tex]\( M = 112.41 \ \text{g/mol} \)[/tex]
2. Convert the cross-section to square meters:
[tex]\[ \sigma = 2.4 \times 10^4 \ \text{barns} = 2.4 \times 10^4 \times 10^{-28} \ \text{m}^2 = 2.4 \times 10^{-24} \ \text{m}^2 \][/tex]
3. Convert the molar mass to kilograms per mole:
[tex]\[ M_{\text{kg/mol}} = 112.41 \ \text{g/mol} \times 10^{-3} \ \text{kg/g} = 0.11241 \ \text{kg/mol} \][/tex]
4. Calculate the number density of cadmium atoms ([tex]\(N\)[/tex]):
[tex]\[ N = \left( \frac{\rho}{M_{\text{kg/mol}}} \right) N_A = \left( \frac{8.6 \times 10^3 \ \text{kg/m}^3}{0.11241 \ \text{kg/mol}} \right) (6.022 \times 10^{23} \ \text{atoms/mol}) \][/tex]
[tex]\[ N \approx 4.612 \times 10^{28} \ \text{atoms/m}^3 \][/tex]
5. Calculate the mean free path ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{1}{N \sigma} = \frac{1}{(4.612 \times 10^{28} \ \text{atoms/m}^3)(2.4 \times 10^{-24} \ \text{m}^2)} \][/tex]
[tex]\[ \lambda \approx 9.043 \times 10^{-6} \ \text{m} \][/tex]
### Part 2: Estimating the Thickness to Remove 99% of Neutrons
The attenuation of neutrons in a material can be described by the exponential attenuation law:
[tex]\[ I(x) = I_0 e^{-x / \lambda} \][/tex]
where [tex]\(I(x)\)[/tex] is the intensity of neutrons after traveling through the material, [tex]\(I_0\)[/tex] is the initial intensity, [tex]\(x\)[/tex] is the thickness, and [tex]\(\lambda\)[/tex] is the mean free path.
To remove 99% of the neutrons:
[tex]\[ \frac{I(x)}{I_0} = 1 - 0.99 = 0.01 \][/tex]
Taking the natural logarithm on both sides, we get:
[tex]\[ \ln\left(\frac{I(x)}{I_0}\right) = \ln(0.01) \][/tex]
[tex]\[ -\frac{x}{\lambda} = \ln(0.01) \][/tex]
[tex]\[ x = -\lambda \ln(0.01) \][/tex]
Since [tex]\(\ln(0.01) = -4.605\)[/tex]:
[tex]\[ x = \lambda \times 4.605 \][/tex]
Substituting the value of [tex]\(\lambda\)[/tex]:
[tex]\[ x = 9.043 \times 10^{-6} \ \text{m} \times 4.605 \][/tex]
[tex]\[ x \approx 4.165 \times 10^{-5} \ \text{m} \][/tex]
### Final Answer:
- The mean free path of thermal neutrons in cadmium is approximately [tex]\(9.044 \times 10^{-6} \ \text{m}\)[/tex].
- The thickness of cadmium required to remove 99% of the neutrons from the beam is approximately [tex]\(4.165 \times 10^{-5} \ \text{m}\)[/tex] or 41.65 micrometers.
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