A narrow beam of alpha particles falls normally on a silver foil behind which a counter is set to register the scattered particles. On substraction of platinum foil of the same mass thickness for the silver foil, the number of alpha particles registered per unit time increased `eta= 1.52 time s`. Find the atomic number of platinum, assuming the atomic number of silver and the atomic masses of both platinum and silver to be known.

Explanation

From the formulas (6.1b) of the book, we find
`(dN_(pt))/(dN_(Ag))=(n_(pt))/(n_(Ag)).(Z_(pt)^(2))/(Z_(Ag)^(2))=eta`
But since the foils have the same mass thickness `(=pd)`, we have
`(n_(pt))/(n_(Ag))=(A_(Ag))/(A_(pt))`
See the problem (6.9) Hence
`Z_(pt)=Z_(Ag).sqrt((etaA_(Ag))/(A_(pt)))`
Substituting `Z_(Ag)= 47,A_(Ag)= 108mA_(pt)= 195` and `eta= 1.52` we get
`Z_(pt)= 77.86~~78`