Estimate distance of neutrino

  • #1

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!
 

Answers and Replies

  • #2
BvU
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What happened to the ##10^{-12}## ?
 
  • #4
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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:
 
  • #5
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?
 
  • #6
BvU
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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
 
Last edited:

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