The rate at which radon is being created is the rate at which radium is decaying, namely A1 N1 or A1 Noe-Ait. But the radon is also decaying at the rate A,N2. Hence, we have dN2 = \1N1 – A2N2, dt or dN2 + A2N2 = A1N1 = 11 Noe¬A1t_ dt This equation is of the form (3.1), and we solve it as follows: I = / A2 dt = Azt, (3.10) N2e^2t A, Noe-Aite^2t dt + c = e(A2-A1)t dt + c= d1 No (A2-A1)t + c, A2 - A1 12, see Problem 19.) Since N2 = 0 at t = 0 (we if A1 + d2. (For the case A1 assumed pure Ra at t = 0), we must have d1 No A1 No 0 = d2 – di or c = Substituting this value of c into (3.10) and solving for N2, we get d, No (e- d2 - A1 - e-Aat). Ait N2 > Example 2. Radium decays to radon which decays to polonium. If at t = 0, a sample is pure radium, how much radon does it contain at time t? Let No = number of radium atoms at t = 0, N1 = number of radium atoms at time t, N2 = number of radon atoms at time t, A1 and A2 = decay constants for Ra and Rn. As in Section 2, we have for radium dN1 -A1 N1, N1 = Noe-^t. dt

The rate at which radon is being created is the rate at which radium is decaying, namely A1 N1 or A1 Noe-Ait. But the radon is also decaying at the rate A,N2. Hence, we have dN2 = \1N1 – A2N2, dt or dN2 + A2N2 = A1N1 = 11 Noe¬A1t_ dt This equation is of the form (3.1), and we solve it as follows: I = / A2 dt = Azt, (3.10) N2e^2t A, Noe-Aite^2t dt + c = e(A2-A1)t dt + c= d1 No (A2-A1)t + c, A2 - A1 12, see Problem 19.) Since N2 = 0 at t = 0 (we if A1 + d2. (For the case A1 assumed pure Ra at t = 0), we must have d1 No A1 No 0 = d2 – di or c = Substituting this value of c into (3.10) and solving for N2, we get d, No (e- d2 - A1 - e-Aat). Ait N2 > Example 2. Radium decays to radon which decays to polonium. If at t = 0, a sample is pure radium, how much radon does it contain at time t? Let No = number of radium atoms at t = 0, N1 = number of radium atoms at time t, N2 = number of radon atoms at time t, A1 and A2 = decay constants for Ra and Rn. As in Section 2, we have for radium dN1 -A1 N1, N1 = Noe-^t. dt

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