Question

A narrow beam of alpha particles with kinetic energy `T= 0.50MeV` and intensity `I= 5.0.10^(5)` particles per second falls normally on a golden foil. Find the thickness of the if at a distance `r=15 cm` from a scattering section of that foil the flux density of scatteered particles at the angle `theta = 60^(@)` to the incident beam is equal to `J= 40` particles`//(cm^(2).s)`.

Answer 1

A scattered flux density of `J` (particles per unit area per second) equals `J//(1)/(r^(2)=r^(2)J` particles scattered per unit time per steradian in the given direction Let `n=` concentration of the gold nuclei in the foil Then
`n=(N_(A)rho)/(A_(Au)`
and the number of `Au` nuclei per unit area of the foil is `nd` where d= thickenss of the foil. Then from Eqn (6.1b) (note `n rarr nd` here)
`r^(2)J=(dN)=md1((Ze^(2))/((4piepsilon_(0))2T))^(2) "cosec"^(4)(theta)/(2)`
Hetre `I` is the number of `alpha`-particle falling on the foil per second
Hence `d=(4T^(2)r^(2)J)/(nI((Ze^(2))/(4piepsilon_(0)))^(2)) sin^(4)theta//2`
Using `Z= 79,A_(Au)= 197,rho= 19.3xx10^(3)kg//m^(3)N_(A)= 6.023xx10^(26)//` kilo mole and other data from the problem we get
`d= 1.47 mu m`