How Many Light Years Must a Neutrino Travel for Observable Interactions?

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The discussion revolves around calculating the distance a neutrino must travel for observable interactions. A user calculated the volume of a cylinder's cross-section and the number of interactions per light year, arriving at approximately 6.626 x 10^10 interactions. The challenge lies in determining the number of light years required for a neutrino to interact, considering that only one in 10^12 neutrinos actually interacts, which significantly reduces the effective volume. Participants suggest using the formula involving Avogadro's number and the interaction probability to derive the necessary distance in light years. The conversation emphasizes the complexity of neutrino interactions and the mathematical approach needed for accurate calculations.
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Homework Statement


https://gm1.ggpht.com/EOgEamclEBUKwVrnxKngFGdqUbOp0GW0TtjQoScZXstPwArYvPID2fM_YB2D_XJWKH1Ci-jnUNqegAMdVPP8yCTfqzo7yoWY0GLRSpZQGnFgJEAfCSxTp7iBJSOGLU0T6ibIDUjyh8eR54LKZgQGAxuBg-gKsocOW2zow-W4wffzBfSHzfAtHEipIDJqV700k6c4OqZG7HH_d4V9sJ_ioT8ddOgdBcSqVSCG7ZYxd8drNw-ylQ-MsE75HXv5a6jpjIJTJ0VynN2M4v6YU8Y05mfNnNaOngEIXxgBue90U3E9A1ANn7ZOMVC9w_GlFMqi4jsmypzV91FOJ4ucU-N5NM8EoWifYczWvV9jC1nXJB3pMNZJ7OFA1YaZTyLlvEQ8xeVn_fptYT8_7EE720PbKMRO85Rqoz_XP-1_ZiNpotXKk5hf-cB_5flXYR4BLAM2Lcl1Sz3D9b80AgFZCtE4Oe5TSE8nLPy6mzIgN7e92uqQayUvyLMu_2H2lBu7xqL8nahwYh8q3kFy7d14iKyAYpg0KGGCOfP_i3P6FUtg70zP_xRSkRS36970vC2zvgQFhPUkLnXIuncXYjUgssSiILhwIl8Vjb-AUjPntefKrgDv3K_vdEhpeTLvi10=w1256-h483-l75-ft

Homework Equations

The Attempt at a Solution


I calculated the volume of a cylinder's cross section and one light year: 10^-36 m^2 (9.46 x 10^15 m) = 9.46 x 10^-21 m^3 = 9.46 x 10^-15 cm^3.

Then I calculated the number of interactions in that cylinder: 6.02 x 10^23 x 11 (9.46 x 10^-15 cm^3) = 6.626 x 10^10 interactions per light year.

How do I calculate the light years a neutrino would have to travel?

Could someone show me the process?

Thanks!
 
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What happened to the ##10^{-12}## ?
 
BvU said:
What happened to the ##10^{-12}## ?
What 10^−12?
 
Only one in ##10^{12}## neutrinos actually interact. this reduces the ##10^{-36}## to ##10^{-48}##.
Fortunately there are three quarks inside every nucleon ... :smile:
 
BvU said:
Only one in ##10^{12}## neutrinos actually interact. this reduces the ##10^{-36}## to ##10^{-48}##.
Fortunately there are three quarks inside every nucleon ... :smile:
Thanks. I'm still not sure how to calculate the number of light years?
 
You have ##N_A * 11 * 3 * 10^{-48} * 10^{6}## area in 1 cm3. I suppose you want 0.5 cm2 so you need a lot of cm. How many ? Divide one by the other and convert to light-years.

Let me know what comes out and if that's in the right ball-park
 
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