An experiment requires minimum beta activity produced at the rate of 346 beta particles per minute. The half- life period of `_(42)Mo^(99)` , which is a beta emitter , is 66.6 h . Find the minimum amount of `_(42)Mo^(99)` required to carry out the experiment in 6.909 h.

Explanation

No. of `beta` particles per minute=346
Time of experiment=6.909 hr
No. of `beta` particles required for the whole experiment=`346xx6.909xx60=143430`
Number of moles `beta-"particiles"` needed for the
experiment=`(143430)/("Avogadro s no")=(143430)/(6.023xx10""^(23))`
=`2.3814xx10""^(-19)"mole"` Half life period of `""_(43)""^(99)Mo=66.6hr=66.6hr`
Distingeration constant or Decay constant
`lambda=(0.693)/(t""_(1//2))=(0.693)/(66.6)hr""^(-1)`
let the initial conc. of `""_(42)Mo""^(99)` needed=ag=No conc. left after the experiment =a-x-N but `x=2.3814xx10""^(-19)mol=143430"atom"` apply radioactive decay equation
`lamda=(2.303)/(t)"log"(a)/(a-x)=(2.303)/(t)"log"(N""_(0))/(N)`
`=(0.693)/(66.6)(2.303)/(6.909)"log"=(a)/(a-143430)`
`"log"=(a)/(a-143430)=(6.909xx0.693)/(2.303xx66.6)=0.0312`
`or "log"=(a)/(a-2.3814xx10""^(-19))=0.0312`
Taking antilogarithm `(a)/(a-2.3814xx10""^(-19))=1.07`
`a=1.07(a-2.3814xx10""^(-19))`
`0.07a=2.518xx10""^(-19)mole`
`a=3.597xx10""^(18)"mole"`
At. wt. of Mpo=99
Amount of Mo=`3.597xx10""^(-18)xx99 =3.56xx10""^(16)g