Find the probability that a muon is passing through your body this instant

JFuld
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Homework Statement



the flux of muons on the Earth's surface is about 100 muons per square meter per second. Estimate the probability that a muon is passing
through your body this instant to within a factor of three

attempt:

let F = flux of muons on Earth's surface = 100/m^2/s

let the average person occupy an area A on the Earth's surface.

then F*A =# of muons penetrating person every second

I don't really know where to go from here. if there are N "instants" in a second then FA/N gives the average # of muons passing through you in an instant?

I feel i am going about this wrong, any help is appreciated.
 
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Hi JFuld! :smile:
JFuld said:
Estimate the probability that a muon is passing
through your body this instant to within a factor of three

… if there are N "instants" in a second

no, an instant is zero time,

it means how many are likely to be inside your body at any fixed time

(so you'll need an estimate of their speed :wink:)
 
ahh thank you, I knew I had no idea what I was doing haha. Well here is what I did in sight of your comment:

Im still using F*A =# of muons penetrating person every second, and I am letting v=vertical velocity of muons. Also, I am picturing the incident muons as a steady stream.

v*(1 second) = d, the distance traveled by muons in one second. So a vertical segment of length d contains 100 muons/m^2.

and A*d corresponds to the # of muons contained in segment d over an area A.

then taking A to equal .25m^2, the volume d*A contains 25 muons at an instant.

and 25/d =# muons per area A, then h*25/d = N = the # of muons occupying a person at an instant

also, my estimate for the muon velocity is: 1*10^8 m/s < v < c

then the range for d is: 1*10^8 m < d < 3*10^8 m

this give the range for N: 4.5*10^-7 < N < 1.5*10^-7 (i aproximate h to be 1.8 m)
 
looks good! :smile:
 
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