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

To carry out experiment,
Rate of `beta-` emission required `=346` particles `mi n(-1)`
Rate `=k.N`
or desired number of atoms to carry out experiment after `6.909hr`
`=(Rate)/(I)=(346xx66.6xx60)/(0.693)=1.995xx10^(6) at oms`
Now, when `N=1.995xx10^(6)` at oms of `Mo` at `t=6.909hr`,
`N_(0)` can be evaluated as `:`
`t=(2.303)/(k)"log" (N_(0))/(N)`
`6.909=(2.303xx66.6)/(0.693)"log" (N_(0))/(N)`
`:. (N_(0))/(N)=1.0745`
`:. N_(0)=Nxx1.0745`
`=1.995xx10^(6)xx1.0745`
`=2.1436xx10^(6) ` atoms of `Mo^(99)`
Weight of `Mo` requried to carry out experiment in `6.909 hr`
`=(2.1436xx10^(6)xx99)/(6.023xx10^(23))g` ltbr. `=3.56xx10^9-16)g`