Activity of a certain radioactive sample is 25 `muCi` at the end of 20 days and 6.25 `mu Ci` at the end of 40 days. What will be its activity at the end of 30 days?

Explanation

We know that `A=A_(0)e^(lambda t)=(A_(0))/(2^(n)), n-(t)/(T)`
given, `25=(A_(0))/(2^((20//T))`
and `6.25=(A_(0))/(2^((40//T))`
`(1)div(2),` we get
`2^(2)=2^([40//T-20//T])`
`"i.e., "(40)/(T)-(20)/(T)=2`
`(20)/(2)=T`
`therefore " "T=10` days
Initial activity`=25=(A_(0)/(2^(20//10)))`
`"i.e., "A_(0)=25xx4=100 mu Ci`
`therefore` Activity of the sample at the end of 30 days will be,
`A=(10)/(2^(30//10))=(110)/(8)=12.5 mu Ci`.