Question

What fraction of 5 MeV α particles will be scattered through angles greater than 4.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thickness 10-8 m?

Answer 1

The fraction of particles is
`6.21 * 10^(-4)`

Step-by-step explanation:

Given :

Alpha particle atomic no.
`Z_(1) = 2`

Gold foil atomic number
`Z_(2) = 79`

Thickness
`t = 10^(-8)` m

Density
`\rho = 19.3`
`(g)/(cm^(3) )`

Kinetic energy of alpha particle
`K = 5 * 1.6 * 10^(-19) * 10^(6)` V

`K = 8 * 10^(-13)` V

Scattered angle
`\alpha =` 4.5°

Fraction of incident particle scattered at angle
`\alpha` is,

`f(\alpha ) = \pi nt ((Z_(1)e Z_(2) e )/(8\pi \epsilon _(o) K ) )^(2) \cot ^(2) ((\alpha )/(2) )`

Where
`n =` number density of particle,
`\epsilon _(o) = 8.85 * 10^(-12)`

For calculating value of
`n`

`n = (\rho N_(A) )/(M)`

Where
`N_(A) = 6.022 * 10^(23)`,
`M = 197` ( Mass number of gold foil )

`n = 5.9 * 10^(28)`
`(atom)/(m^(3) )`

Put the all value in above equation,

`f(4.5 ) = 3.14 * 5.9 * 10^(28) * 10^(-8) ((2 * 79 (1.6 * 10^(-19) ) ^(2) )/(8* 3.14 * 8.85 * 10^(-12) 8 * 10^(-13) ) )^(2) \cot ^(2) ((4.8 )/(2) )`

`f (4.5 ) = 0.96 * 647.78 * 10^(-6)`

`f (4.5) = 6.21 * 10^(-4)`
`(atoms)/(m^(2) )`

Therefore, the fraction of particles is
`6.21 * 10^(-4)`