Atomicn number of of silver metal is 47. calculate the speed at which a beam of protons has to be fired at a sheet of silver foil if the protons were able to approach to whithin `2.5xx10^(-14)` m of the silver nucleus?

Explanation

The distance between protons and slilver nucleas is
`r=2.5xx10^(-14)m`
Mass of proton is `m,=1.67xx10^(-27 ) kg`
Charge of protons is `q_(!)=1.6xx10^(-19)C`
Atomic number of silver , Z=47
Charge of silver nueleus, `q_(2)=Ze=(47)(1.6xx10^(-19)C)`
By energy conservation at a distance r the kinetic energy of proton must be equal to the potential energy of proton at that point. Thererfore we get
`K=UrArr(1)/(2)mv^(2)=(kq_(1)q_(2))/(r)`
Thererfore `v=sqrt((2q_(1)q_(2))/(mr))`
`=sqrt((2xx(9xx10^(9))(1.6xx10^(-19))(47)(1.6xx10^(-19)))/((1.67xx10^(-2))(2.5xx10^(-14))))=2.28xx10^(7)ms^(-1)`