Nuclear physics - solar neutrinos interacting

[Benlaww]

Homework Statement

A tank of 12C2 37Cl4, with density 1622 kg m−3, is used as a neutrino detector to measure the flux of neutrinos produced in the Sun. A neutrino has a cross-section of interacting with a 37Cl nucleus of approximately 12 fb (1 femtobarn equals 10−43 m^2).

Write down the nuclear decay that occurs when a solar neutrino interacts with a 37Cl nucleus (try rearranging beta-decay equations).

What is the probability that the neutrino will interact when travelling through 10m of detector? (think of the definition of cross-section).

Relevant Equations

beta plus : p -> n + e- + v
beta minus: n -> p + e+ + anti-v

For the first part I thought you'd have either, p + anti-v -> n + e+ and n + v -> p + e-, but I thought it'd probably be the latter as it's a 'normal' neutrino not an anti neutrino? But do I need to include the actual elements in the equation?

For the second part I have multiplied the density by the cross section to get 1.9*10^-39 kgm-1 but I don't know if this is right or where to go next.

 

[haruspex]

Can't help with the first part, but for the second...
Consider a pipe through the tank with a 1m² cross section.
How many Cl atoms are in the pipe?
If you look along the pipe from one end, and assume you can see all the Cl atoms without overlap, what fraction of the cross section do they cover?

 

[Steve4Physics]

Homework Statement:: A tank of 12C2 37Cl4, with density 1622 kg m−3, is used as a neutrino detector to measure the flux of neutrinos produced in the Sun. A neutrino has a cross-section of interacting with a 37Cl nucleus of approximately 12 fb (1 femtobarn equals 10−43 m^2).

Write down the nuclear decay that occurs when a solar neutrino interacts with a 37Cl nucleus (try rearranging beta-decay equations).

What is the probability that the neutrino will interact when traveling through 10m of detector? (think of the definition of cross-section).
Relevant Equations:: beta plus : p -> n + e- + v
beta minus: n -> p + e+ + anti-v

For the first part I thought you'd have either, p + anti-v -> n + e+ and n + v -> p + e-, but I thought it'd probably be the latter as it's a 'normal' neutrino not an anti neutrino? But do I need to include the actual elements in the equation?

For the second part I have multiplied the density by the cross section to get 1.9*10^-39 kgm-1 but I don't know if this is right or where to go next.

For part 1, the question refers to neutrinos, not antineutrinos. And Cl-37 is specifically named. So it looks like they are asking for a balanced equation starting:
$\nu + {^{37}Cl} →$
(I would describe this as a ‘reaction’ rather than a ‘decay’.)

For part 2, I agree with @haruspex.

Edt - typo'.

 

[Benlaww]

For part 1, the question refers to neutrinos, not antineutrinos. And Cl-37 is specifically named. So it looks like they are asking for a balanced equation starting:
$\nu + {^{37}Cl} →$
(I would describe this as a ‘reaction’ rather than a ‘decay’.)

For part 2, I agree with @haruspex.

Edt - typo'.

Would the element it decays to just be an isotope of Cl or am I supposed to be able to tell which new element it would be?

 

[phyzguy]

Would the element it decays to just be an isotope of Cl or am I supposed to be able to tell which new element it would be?

Once you've figured out the reaction, you're supposed to be able to deduce whether the nucleus is still Cl or whether it is converted to something else.

 

[Benlaww]

Can't help with the first part, but for the second...
Consider a pipe through the tank with a 1m² cross section.
How many Cl atoms are in the pipe?
If you look along the pipe from one end, and assume you can see all the Cl atoms without overlap, what fraction of the cross section do they cover?

I'm still a bit confused with working this out, I have got the volume of the 'pipe' as 10 m^2 and so the density times volume gives 16220 kg and then I've divide this mass by the mass of 1 Cl atom to give 2.64*10^-26 atoms?..

 

[Steve4Physics]

Hi @Benlaww. I can see you are struggling. Here is a guide for some of the basic steps you need to go through.

For part 1 you need to decide what has happened to the number of protons and neutrons inside the Cl-37 nucleus. Then (with reference to the periodic table if needed) you need to work out what the changed nucleus is. Here are some basic questions/steps:
Q1 how many protons in Cl-37?
Q2 how many neutrons in Cl-37?
Q3 how many protons in the changed nucleus?
Q4 how many neutrons in the changed nucleus?
Q5 what is the new nucleus?

For part 2 you imagine a 10m long cylinder full of C₂Cl₄. You can take the cross-sectional area as 1m² for convenience, or call it A (then A will cancel out later).
Q6 what is the volume of the cylinder?
Q7 what is the mass of C₂Cl₄ in the cylinder?
Q8 what is the molecular weight of C₂Cl₄?
Q9 how many moles of C₂Cl₄ are present in the cylinder?
Q10 how many molecules of C₂Cl₄ are present in the cylinder?
Q11 how many atoms of Cl are present in the cylinder?

Then you're nearly home.

Edited - typo's corrrected.

 

[Benlaww]

Hi @Benlaww. I can see you are struggling. Here is a guide for some of the basic steps you need to go through.

For part 1 you need to decide what has happened to the number of protons and neutrons inside the Cl-37 nucleus. Then (with reference to the periodic table if needed) you need to work out what the changed nucleus is. Here are some basic questions/steps:
Q1 how many protons in Cl-37?
Q2 how many neutrons in Cl-37?
Q3 how many protons in the changed nucleus?
Q4 how many neutrons in the changed nucleus?
Q5 what is the new nucleus?

For part 2 you imagine a 10m long cylinder full of C₂Cl₄. You can take the cross-sectional area as 1m² for convenience, or call it A (then A will cancel out later).
Q6 what is the volume of the cylinder?
Q7 what is the mass of C₂Cl₄ in the cylinder?
Q8 what is the molecular weight of C₂Cl₄?
Q9 how many moles of C₂Cl₄ are present in the cylinder?
Q10 how many molecules of C₂Cl₄ are present in the cylinder?
Q11 how many atoms of Cl are present in the cylinder?

Then you're nearly home.

Edited - typo's corrrected.

Thank you, I've got that I have 5.89 * 10^28 molecules of C2Cl4 but I'm still a bit confused on how it all relates.

 

[Steve4Physics]

... I have 5.89 * 10^28 molecules of C2Cl4 ...

I disagree with your value for the number of molecules slightly. If you go by what the question says, all the chlorine is Cl-37. No other isotopes of chlorine are present. You should take the atomic weight of Cl-37 to be 37grams. The molecular mass of C₂Cl₄ is then 2*12 + 4*37 = 172grams = 0.172kg. That's the best value to use when calculating the number of molecules.

(In real life the chlorine atoms will be a mixture of mainly Cl-35 and Cl-37. But, from the wording in the question, you are meant to take all the chlorine to be Cl-37 in this problem. That changes the molcular weight you need to use for the C₂Cl₄.)

After this correction, you have successfully answered Q10 (see post #7). What is the answer to Q11?

Then, thinking about what @haruspex said in post #2, what is the total area of the Cl-37 nuclei as ‘seen’ by the incident neutrinos? (Remember each nucleus presents an area of 12fb).

Did you sort out part 1 of the question?

 

[Benlaww]

I disagree with your value for the number of molecules slightly. If you go by what the question says, all the chlorine is Cl-37. No other isotopes of chlorine are present. You should take the atomic weight of Cl-37 to be 37grams. The molecular mass of C₂Cl₄ is then 2*12 + 4*37 = 172grams = 0.172kg. That's the best value to use when calculating the number of molecules.

(In real life the chlorine atoms will be a mixture of mainly Cl-35 and Cl-37. But, from the wording in the question, you are meant to take all the chlorine to be Cl-37 in this problem. That changes the molcular weight you need to use for the C₂Cl₄.)

After this correction, you have successfully answered Q10 (see post #7). What is the answer to Q11?

Then, thinking about what @haruspex said in post #2, what is the total area of the Cl-37 nuclei as ‘seen’ by the incident neutrinos? (Remember each nucleus presents an area of 12fb).

Did you sort out part 1 of the question?

I think I got part 1, I've got that Cl-37 + v --> Ar-37 + e-

Changing the molecular mass to 0.172kg, I then get 5.68 * 10^28 molecules so then the number of Cl atoms would be four times this? giving 2.27 * 10^29. So the total area is 2.73 * 10^30 fb.

 

[Steve4Physics]

I think I got part 1, I've got that Cl-37 + v --> Ar-37 + e-

Changing the molecular mass to 0.172kg, I then get 5.68 * 10^28 molecules so then the number of Cl atoms would be four times this? giving 2.27 * 10^29. So the total area is 2.73 * 10^30 fb.

Agreed on all points!

(Note, each molecule contains four Cl-37 atoms. So there is no 'trick' - you are just multiplying the number of molecules by four to get the number of Cl-37 atoms.)

Consider this simple problem. A sheet of paper has area 100cm². Three coins, each with area 2cm², are placed (non-overlapping) on the paper. A drop of rain fall towards the paper. What is the probability that the rainsdrop hits a coin?

If you can do that, the neutrino calculation is essentially the same. (Of course you will need to work in square metres.)

 

[Benlaww]

Agreed on all points!

(Note, each molecule contains four Cl-37 atoms. So there is no 'trick' - you are just multiplying the number of molecules by four to get the number of Cl-37 atoms.)

Consider this simple problem. A sheet of paper has area 100cm². Three coins, each with area 2cm², are placed (non-overlapping) on the paper. A drop of rain fall towards the paper. What is the probability that the rainsdrop hits a coin?

If you can do that, the neutrino calculation is essentially the same. (Of course you will need to work in square metres.)

Thank you!