A narrow beam of protons with kinetic energy `T=1.4 meV` falls normally on a brass foil whose mass thickness `rho d = 1.5mg//cm^(2)`. The wavelength ratio of copper and zinc in the foil is equal to `7 : 3` respectively. Find the fraction of the protons scattered through the angles exceeding `theta_(0)=30^(@)`.

Explanation

The relevant fraction can be immediately written down (Note that the particles are protons)
`(DeltaN)/(N)=((e^(2))/((4pi epsilon_(0))2T))^(2) pi "cot"^(2)(theta_(0))/(2)(n_(1)Z_(1)^(2)+n_(2)Z_(2)^(2))`
Here `n_(1)(n_(2))` is the number of `Z_(n)(Cu)`nuclei per `cm^(2)` of the foil and `Z_(1)(Z_(2))` is the atomic number of `Z_(n)(Cu)`. Now
`n_(1)=(rhodN_(A))/(M_(1))= 0.7,n_(2)=(rhodN_(A))/(M_(2))=0.3`
Here `M_(1),M_(2)` are the mass numbers of `Z_(n)` and `Cu`.
Then, substituting the values `Z_(1)= 30,Z_(2)= 29,M_(1)= 65.4,M_(2)= 63.5`, we get
`(DeltaN)/(N)= 1/43xx10^(-3)`