A narrow beam of alpha particles with kinetic energy `T== 0.50 MeV` falls normally on a golden foil whose t=mass of thickness is `rho d= 1.5 mg//cm^(2)` The beam intensity is `I_(0)= 5.0.10^(5)` particles per second. Find the number of alpha particles scattered by the foil during the time interval `tau=30` m in into the angular interval:
(a) `59-61^(@)` , (b) over `theta_(0)=60^(@)`.

Explanation

(a) From Eqn. (6.1b) we get
`dN=I_(0)tau(rho.d.N_(A))/(A_(Au))((Ze^(2))/((4 pi epsilon_(0))2T))^(2)(2pi sin theta d theta)/(sin^(4)theta//2)`
(we have used `dOmega= 2 sin theta d theta` and `N= I_(0)I`)
From the data `d theta=2^(@)=(2)/(57.3) radian`
Also `Z_(Au)= 79,A_(Au)= 197`. Putting the value we get
`dN= 1.63xx10^(6)`
(b) This number is
`N(theta_(0))=I_(0)tau((rhodN_(A))/(A_(Au)))((Ze^(2))/((4pi epsilon_(0))2T))^(2) 4piint_(theta_(0))^(x)("cos"(theta)/(2)d theta)/("sin"^(3)(theta)/(2))` The integal is
`2int_("sin"(theta_(0))/(2))^(1)(dx)/(x^(2))=(1)/(2)[(-1)/(2x^(2))]_("sin"(theta_(0))/(2))^(1)="cot"^(2)(theta_(0))/(2)`
Thus `N(theta_(0))= pi nd(Ze^(2)/((4piepsilon_(0)T)))^(2)I_(0)tau"cot"^(2)(theta_(0))/(2)` ltbr where `n` is the concentration of nuclei in the foil. `(n= rhoN_(1)//A_(Au))`
Substitution gives
`N(theta_(0))= 2.02xx10^(7)`