Neutron absorbed by a sheet of indium, % chance

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Homework Help Overview

The problem involves a neutron passing through a thin sheet of indium, specifically 115In, and seeks to determine the probability of absorption given certain physical parameters such as thickness, density, and absorption cross section.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of the neutron's absorption probability, questioning the density conversion and the determination of the number of target nuclei per unit volume. There are attempts to clarify the correct density value and its implications for the calculations.

Discussion Status

The discussion includes various attempts to clarify the density of the material and how it affects the calculation of the number of atoms. Some participants express confusion about the calculations and the correct application of formulas, while others provide corrections and suggestions for re-evaluating certain values.

Contextual Notes

There is an ongoing debate about the correct density value and its conversion from g/cm³ to kg/m³, which affects the calculations. Participants also note that the area of the sheet is not provided, complicating the determination of the neutron's absorption probability.

richphys
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Homework Statement



A neutron is passing through a thin sheet of 115In of thickness 0.01 cm. Given that the density of the sheet is 7.31 g cm-3 and that the absorption cross section is 100 barns, what is the chance of the neutron being absorbed? You may assume the neutron is not scattered.
a) 1.0%
b) 3.8%
c) 5.7%
d) 20%

Homework Equations



Chance of projectile not getting through = ratio (black area / total area) = σT n A Δx / A

The Attempt at a Solution



Δx = 10-4m
σT= 10-26m2
Density =7.31 *10-9 kg m-3
Volume = mass 115ln / density of sheet = 2.61x10-17 m3

Chance = (10-26m2) x (n/2.61x10-17 m3) x 10-4m

A cancels, but I don't know how to get a value for n (target nuclei n per unit volume). I am completely stuck on this question.
 
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richphys said:
Density =7.31 *10-9 kg m-3
This is incorrect. Check the conversion to kg/m3

A cancels, but I don't know how to get a value for n (target nuclei n per unit volume). I am completely stuck on this question.
One way to think about it is

mass of 1 m3 of material = (number of atoms in 1 m3 of material) x (mass of 1 atom)
 
Is density 7.31 x 10-10 kg m-3 ?

For n, is 3.83 x 1045 correct?

I'm still getting nowhere near the answer. Totally stuck.
 
richphys said:
Is density 7.31 x 10-10 kg m-3 ?
No. How would you fill in the question marks in the conversion below?

$$1 \frac{\rm g}{\rm {cm^3}} = 1 \frac{ \rm g}{ \rm {cm^3}} \times \frac{1 \,\, \rm {kg}}{? \,\, \rm g} \times \frac{? \,\, \rm {cm^3}}{1 \,\, \rm m^3}$$
 
TSny said:
No. How would you fill in the question marks in the conversion below?

$$1 \frac{\rm g}{\rm {cm^3}} = 1 \frac{ \rm g}{ \rm {cm^3}} \times \frac{1 \,\, \rm {kg}}{? \,\, \rm g} \times \frac{? \,\, \rm {cm^3}}{1 \,\, \rm m^3}$$

I have density as 7310 kg m-3

Then I did

number of atoms n = mass of 1m3 of material / mass 1 atom
n =(7310 x 10 -30 kg) / (1.908 x 10 -25 kg)

chance = (10-26m2) x (0.0383/10-30 m3) x (10-4m) = 0.0383

I multipied that by 100 and got 3.83% (answer b)

Is this correct?
 
richphys said:
I have density as 7310 kg m-3

Then I did

number of atoms n = mass of 1m3 of material / mass 1 atom
n =(7310 x 10 -30 kg) / (1.908 x 10 -25 kg)

chance = (10-26m2) x (0.0383/10-30 m3) x (10-4m) = 0.0383

I multipied that by 100 and got 3.83% (answer b)

Is this correct?
Yes, except I'm sure you didn't mean to type the factor of 10 -30 when writing
n =(7310 x 10 -30 kg) / (1.908 x 10 -25 kg)
 
TSny said:
Yes, except I'm sure you didn't mean to type the factor of 10 -30 when writing

Yes, I did mass = density x volume which was (7310 kg m-3) x 10-30

Have I somehow got the answer even by doing this step wrong?
 
richphys said:
number of atoms n = mass of 1m3 of material / mass 1 atom
This statement is correct. [EDIT: The left side should actually say "number of atoms in 1 m3 = n" ]
Note that the numerator on the right is the mass of 1 m3 of material.

n =(7310 x 10 -30 kg) / (1.908 x 10 -25 kg)
The numerator should represent the mass of 1 m3 of material.
Earlier, you found correctly that the density of the material is 7300 kg/m3. So, what is the mass of 1 m3 of material.
 
Last edited:
richphys said:
Yes, I did mass = density x volume which was (7310 kg m-3) x 10-30

Have I somehow got the answer even by doing this step wrong?
It was a bit confusing that you wrote
richphys said:
number of atoms n = mass of 1m3 of material / mass 1 atom
You meant
number of atoms n = mass of material / mass 1 atom.
(I gather the area is 10-26m2.)
 
  • #10
haruspex said:
It was a bit confusing that you wrote

You meant
number of atoms n = mass of material / mass 1 atom.
n in the formula in the relevant equations of the OP is the number of atoms per unit volume. I should have noted the error in post #5 where
number of atoms n = mass of 1m3 of material / mass 1 atom
The left side of this equation should read

number of atoms in 1 m3 = n = mass of 1m3 of material / mass 1 atom

Sorry for missing that.

(The area of the sheet of material is not given, so the volume and mass of the sheet are not known.)
 
  • #11
Yeah I don't really understand what's going on and I've run out of time so I will just put the answer as 3.8%. Thanks for the help
 

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