Question

# A radioactive isotope has a half-life of T years. How long will it takethe activity to reduce to a) 3.125%, b) 1% of its original value?

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Solution

## a) Given: The half life of the radioactive isotope is T years and the reduction in the activity is 3.125%. According to the law of radioactive decay, N= N 0 e −λt N N 0 = e −λt Where, N is the remaining number of molecules, N 0 is the initial number of atoms, λ is the decay constant and t is the time. By substituting the given values in the above expression, we get 3.125 100 = e −λt 1 32 = e −λt λt= log e ( 32 ) t= 3.4657 λ We know that, λ= 0.693 T By substituting the given value in the above expression, we get t= 3.466 0.693 T =5T years Thus, the isotope will take about 5T years to reduce to 3.125% of its original value. b) Given: The reduction in the activity is 1%. According to the law of radioactive decay, N= N 0 e −λt N N 0 = e −λt By substituting the given values in the above expression, we get 1 100 = e −λt 1 100 = e −λt λt= log e ( 100 ) t= 4.6052 λ We know that, λ= 0.693 T By substituting the given value in the above expression, we get t= 4.0652 0.693 T =6.65T years Thus, the isotope will take about 6.65T years to reduce to 1% of its original value.

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