The intensity of gamma radiation form a given source is I. On passing through 27 mm of lead, it is reduced to `I//8`. The thickness of lead which will reduce the intesity to `I//2` will be:

Answer

9

Explanation

form `I=I_(0)e^(-mux)`
`-mux=log. (I)/(I_(0))`
`=mu(27)=log. (I_(0))/(8I_(0))=3log.((1)/(2))`
and `-mu(x')=log.((1)/(2))`
Dividing, we get, `27/(x')=3`
`x'=9mm`