Cosmic ray muons have a typical energy of 6 GeV. As they pass through the atmosphere, which you may assume to be pure nitrogen, Bremsstrahlung radiation will be produced. Predict the Bremsstrahlung spectrum that will be observed.

205.9 nm, ultraviolet spectrum

Explanation: Bremsstrahlung is simply the abrupt stop of an accelerated electron by another atom. The abrupt stop (by the atom) will cause the high energetic electron to knock off electron off the atom which is given out as electromagnetic radiation.

The mathematical equation that defines Bremsstrahlung is given below as

E =hc/λ

Where E = energy of the electron (in eV) = 6ev = 9.654×10^-19 J

h = planck constant = 6.626×10^-34 Js

c = speed of light = 3×10^8 m/s

λ = wavelength of radiation

By substituting parameters, we have that

9.654×10^-19 = 6.626×10^-34 × 3×10^8/ λ

λ = 6.626×10^-34 × 3×10^8/ 9.654×10^-19

λ = 0.0000002059 = 205.9 nm

This is the minimum wavelength of radiation and judging from the electromagnetic spectrum the Bremsstrahlung spectrum will be on the ultra violet spectrum (10 nm to 400 nm)

the one with the highest speed

answer: (431,2)

step-by-step explanation: