Neutron absorbed by a sheet of indium, % chance

[richphys]

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.

 

[TSny]

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)

 

[richphys]

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.

 

[TSny]

Is density 7.31 x 10^-10 kg m^-3 ?

No. How would you fill in the question marks in the conversion below?

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

 

[richphys]

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?

 

[TSny]

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)

 

[richphys]

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?

 

[TSny]

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.

 

[haruspex]

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

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.)

 

[TSny]

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.)

 

[richphys]

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