`._(92)U^(238)` is radioactive and it emits `alpha` and `beta` particles to form `._(82)Pb^(206).` Calculate the number of `alpha` and `beta` particles emitted in this conversion.
An ore of `._(92)U^(238)` is found to contain `._(92)U^(238)` and `._(82)Pb^(206)` in the weight ratio of `1: 0.1.` The half- life -period of `._(92)U^(238)` is `4.5xx 10^(9) yr.` Calculate the age of the ore.

Explanation

`"Let the no. of " alpha-"particals emited"=x`
`"Let the no. of "beta-"particals emitted"=y`
`"Equation of radioactive decay"`
`""_(92)""^(238)U=""_(82)""^(206)Pb +xalpha+ybeta`
`""_(92)""^(238)U=""_(82)""^(206)Pb+x""_(2)He""^(4)+y""_(-1)e""^(0)`
`"Equating atomic mass 238 = 206" +4x`
`x=8 ("No. of "alpha-"particals")`
`"Equating atomic number"`
`92=82+8xx2-y`
`ory=98-92=6`
`or"ore"to""_(82)Pb""^(206)+""_(92)U""^(238)`
`" "1:1" "`
`" "0.1" "0.1" "`
`"One mole of """_(82)Pb""^(206)" is ontained from one mole of"`
`""_(92)U""^(238)" "0.1" mole of " ""_(82)Pb""^(208)=0.1" mole of " ""_(92)U""^(238)`
`"Let the amount of """_(92)U""^(238)=1g`
`""_(82)Pb""^(206)=""_(92)U""^(238)`
`"206g of Pb is obtained from """_(92)U""^(238)`
`"Let "0.1" "g" of Pb is obtained from"`
`U""^(238)=(238)/(206)xx0.1=0.1155g`
`"Initial conc. no"=1+0.1155=1.1155g`
`"Final conc. no."=1.0`
`t""_(1//2)=4.5xx10""^(9)yr" of " ""_(92)U""^(238)`
`lambda=(0.693)/(t""_(1//2))=(0.693)/(4.5xx10""^(9)`
`"Let t be the age of ore"`
`t=(2.303)/(lambda)log.(N""_(0))/(N)`
`=(2.303xx4.5xx10""^(9))/(0.693)log.(1.1155)/(1.00)`
`t=(2.303xx4.5xx10""^(9))/(0.693)log1.1155`
`"Age of ore"=t=7.09xx10""^(8)" years"`