The mean free path of neutrinos through lead is around one light-year. So, taking the thickness of the body to be 1/2 a meter, you would expect the probability of any individual neutrino to interact with the body to be ~5 x 10^-17. So you'd ballpark have around a 20--40% chance that a single neutrino interacts with your body. It would probably cause a localized radiation burn. Detectable, but probably not lethal unless you got really unlucky with where it hit you.

The mean free path of much lower energy neutrinos in lead is about a light year.

The MFP of a 120 PeV neutrino in lead would be something like 10 kilometers, I think.

More like 100 km I'd think but yeah, the neutrino nucleon cross section gets much bigger at high energies

Total energy of impact would be 120 PeV x 10^16 = 120 x 10^31 eV = ~60 kilotons TNT, or 4 Hiroshimas.

So BIG boom.

Since the velocity is so close to the speed of light, you can think of this like the energy released by annihilating a ping pong ball made of antimatter.

Edit: Commenter asked what would happen if they "hit", so I'm assuming a hypothetical 100% collision. But yes to stop 1/e of a neutrino beam with normal matter, you'd need a light year of lead.

Sounds like a great topic for an xkcd video

There is a what if about it. https://what-if.xkcd.com/73/

> big boom

The probability of interaction of neutrinos with matter increases with the energy. I've asked o1 to estimate the mean free path of a 120 PeV neutrino in water and it came up with 1000km. So let's say, conservatively, that 10^-7 of the total energy gets deposited in your body when the beam goes through. The mass equivalent of a ping pong ball is about 2.5x10^14 J, which gives us 2.5x10^7 J total, or about 6kg TNT equivalent. This is only an order-of-magnitude estimate, but it would definitely not be healthy.