A radioisotope has a half-life of 10 h . Determine the total number of disintegrations that occur in the tenth hour, measured from a time when the activity was 1 Ci .

Step-by-step solution

Initial activity A₀ = 1 Ci = 3.7 10¹⁰ disintegrations/s . Half-life T_ 1/2 = 10 h = 36000 s . Decay constant = 2 T_ 1/2 = 2 10 h ⁻¹ . The tenth hour corresponds to the time interval from t = 9 h to t = 10 h . The total number of disintegrations N in this interval is given by integrating the activity A(t) : N = ₉¹⁰ A(t) dt = ₉¹⁰ A₀ e^ - t dt N = A₀ ( e^ -9 - e^ -10 ) Since e^ - t = 2^ -t / T_ 1/2 , we can rewrite this as: N = A₀ T_ 1/2 2 ( 2^ -9/10 - 2^ -10/10 ) Substituting the values (with T_ 1/2 in seconds to mat