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Neutrinos and their detection

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Homework Statement:

3. SN 1987A, which occurred in the LMC (50 kpc away), is thought to have produced 1058 neutrinos. Kamiokande II, which had 1000 detectors, each with a radius of 0.25 m, detected 12 of those neutrinos.

(a) What is the total area of a sphere extending from the LMC to Earth?

(b) What was SN 1987A’s neutrino flux at the Earth? (Flux is defined as number of neutrinos per area = neutrinos per square meter.)

(c) What is the total collecting area of the 1000 Kamiokande II detectors?

(d) How many SN 1987A neutrinos passed through the Kamiokande II detectors?

(e) What was the efficiency of Kamiokande II? (Fraction of the neutrinos entering the detectors actually detected.)

Relevant Equations:

A = 4pi(r)^2
V= 4/3 * pi(r)^3
Assuming that this sphere has a radius of 50kpc, I've converted to m (1.543e21) and plugged into the area equation for a total area of 2.992e43 m^2. From here I've talked myself into circles, and I honestly don't know where to go next. Any help or guidance would be greatly appreciated!
 
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Answers and Replies

  • #2
haruspex
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total area of 7.48e42 m^2
So with 10e58 neutrinos total, how many per sq m at the earth?
 
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So with 10e58 neutrinos total, how many per sq m at the earth?
I know it's relatively simple, but that's part of what I keep getting tripped up about. Can I set up a proportion with 10e58/Area of sphere = x/Area of Earth?
 
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haruspex
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I know it's relatively simple, but that's part of what I keep getting tripped up about. Can I set up a proportion with 10e58/Area of sphere = x/Area of Earth?
Almost, but what exactly do you mean by area of earth here?
 
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Almost, but what exactly do you mean by area of earth here?
A poor choice of words, would it be better to say the size of the earth, as in its diameter?
 
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haruspex
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A poor choice of words, would it be better to say the size of the earth, as in its diameter?
No, it's an area, but what area exactly?
Think of it from the perspective of the approaching neutrinos. What does the earth look like as a target?
 
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No, it's an area, but what area exactly?
Think of it from the perspective of the approaching neutrinos. What does the earth look like as a target?
Earth would look like a point?
 
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haruspex
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Earth would look like a point?
Yes, but that's not helpful. Magnify the point; what does it look like now?
 
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Yes, but that's not helpful. Magnify the point; what does it look like now?
The area of a circle?
 
  • #10
haruspex
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The area of a circle?
Right. That's the area you need, not the whole surface area of the earth.
 
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Right. That's the area you need, not the whole surface area of the earth.
Thank you! So I've gotten the flux to be 4.27e28, does that sound right? It feels quite big.

And from there I can multiply 1000 by (pi(.25)^2) to find the answer to c? (196.35 m^2)
 
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haruspex
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Thank you! So I've gotten the flux to be 4.27e28, does that sound right? It feels quite big.

And from there I can multiply 1000 by (pi(.25)^2) to find the answer to c? (196.35 m^2)
Sorry, just reread the question. It doesn't ask for the total of neutrinos through the earth. It asks for the neutrinos per sq m at the earth.
So it is 10e58/Area of sphere = x/1m^2.
 
  • #13
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Sorry, just reread the question. It doesn't ask for the total of neutrinos through the earth. It asks for the neutrinos per sq m at the earth.
So it is 10e58/Area of sphere = x/1m^2.
So that's just 10^58 divided by the area of the sphere? And that would be the flux? And how then would I approach d and e?
 
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haruspex
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So that's just 10^58 divided by the area of the sphere? And that would be the flux? And how then would I approach d and e?
By answering c first. Ah, you did that...
So you have the neutrinos per sq m, and you have the total sq m of the detectors. (We will have to assume these were directly facing the source.)
So how many would go through them?
 
  • #15
Orodruin
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Please note that this question seriously misrepresents how neutrino detection works in a water Cerenkov detector.
 

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