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` particle `min^(-1)`
Rate `= lambda. N`
or desired no. of atoms to carry out experiment after `6.909 hr`
`= ("rate")/(lambda) = (346xx66.6xx60)/(0.693)`
`= 1.995xx10^(6)` atoms
Now,
when `N = 1.995xx10^(6)` atoms of `Mo` at `t = 6.909` hr
`N_(0)` can be evaluated as `t = (2.303)/(lambda) log_(10) (N_(0))/(N)`
`6.909=(2.303xx66.6)/(0.693)log_(10)(N_(0))/(N)`
`:. (N_(0))/(N) = 1.0745`
`:. N_(0) = N xx 1.0745 = 1.995xx10^(6) xx 1.0745`
`= 2.1436xx10^(6)` atom of `Mo^(99)`
`:. Wt` of `Mo` required to carry out experiment in `6.909` hours
`= (2.1436xx10^(6)xx99)/(6.023xx10^(23))g = 3.56xx10^(-16)g`