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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? |
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Answer» Solution :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`. |
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