The answer actually depends acutely on the properties of the incoming neutrinos, especially the energy. High energy neutrinos interact more often and release greater amounts of energy after interacting with nucleons, thus increasing the effect of exposure. For the sake of argument, let's limit ourselves to antineutrinos that have an incoming energy of around 10 MeV, similar to solar neutrinos. In this energy regime, antineutrinos interact with protons in hydrogen atoms in water molecules primarily through the inverse beta decay, namely
[math] \bar{\nu}_{e} + p \to e^{+} + n [/math]
and the cross-section of this interaction dominates other stuff. The number of such decays N depend on three things: the cross-section of this interaction [math]\sigma[/math], the neutrino flux [math]\Phi[/math], and the number of protons [math]n_{p}[/math] available for this interaction:
[math] N = \Phi \sigma n_{p} [/math]
I found a very nice plot of the cross-section on Slide 55 of [1] (I think it's from Cowan-Reines), giving a cross-section of [math]\sigma = 10^{-40} \textrm{cm}^2 [/math] for 10 MeV neutrinos. A typical human body contains about [math]10^{28}[/math] protons in hydrogen atoms (the only nucleon that can undergo this interaction) or so. Let's use the solar neutrino flux as a point of comparison, [math] \Phi = 10^{11} \textrm{cm}^{-2} \textrm{s}^{-1}[/math]. That gives about [math]0.01 \textrm{s}^{-1} [/math] incidences of inverse beta decay from incoming neutrinos. That's 1 decay every 100 seconds, somewhere in your body.
Let's conservatively estimate the emitted positron to have about 1 MeV of energy. So your body experiences a radiation dose of about 0.01 MeV/s, which comes up to be about [math]10^{-15} \textrm{J/s}[/math]. That's... puny, coming up to about [math]10^{-14}[/math] Sieverts per hour.
The LD50 for succumbing to radiation poisoning in 60 days is supposedly about 3.5 Sieverts, although there's no mention about how quickly this radiation has to be delivered [2]. Let's assume that this radiation is delivered in about an hour. To reach the LD50/60 levels, you need to increase the neutrino flux by about [math]10^{14}[/math] times, or a hundred trillion times, compared to the current solar neutrino flux. That's a flux of about
[math] \Phi_{kill} = 10^{25} cm^{-2} s^{-1} [/math]
That's still a trillion times larger than the antineutrino flux used by Cowan and Reines to calculate exactly the cross-section we've just discussed, and their experiment was just 11 m from a nuclear reactor using significantly lower energy neutrinos [3]!
So, how might we find someplace in our galaxy where this kind of lethal flux of neutrinos might exist? Well, for example, supernova SN 1987A produced [math]10^{58}[/math] neutrinos at a distance of 50 kpc from us in a burst lasting about 10 seconds. The flux from this is about 3 billion neutrinos / (cm^2 s) on Earth. That's just one order less than the solar neutrino flux, and the average energy is in the same order (~10 MeV) [4]. To get the flux up to LD50 levels, you'll have to get closer by a factor of [math]10^8[/math], which turns out to be about 100 AUs. I think a human will have other problems to worry about at that distance though!
[1] http://ift.tt/1VqE2nK.
[2] How Much Radiation Does it Take to Kill You?
[3] Cowan–Reines neutrino experiment
[4] SN 1987A
Read other answers by Hongwan Liu on Quora:
- What was the initial objective of the CERN experiment in which scientists discovered neutrinos "traveling faster than light"?
- What are some funny neutrino jokes?
- What are the most interesting differential equations in science and mathematics?
from Quora http://ift.tt/25ubOv9
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