Question

Suppose that a radioactive sample contains equal numbers of two radioactive nuclides \(A\) and \(B\) at \(t=0 .\) A has a half-life of \(3.0 \mathrm{h},\) while \(\mathrm{B}\) has a half-life of \(12.0 \mathrm{h}\) Find the ratio of the decay rates or activities \(R_{\mathrm{A}} / R_{\mathrm{B}}\) at (a) \(t=0,\) (b) \(t=12.0 \mathrm{h},\) and \((\mathrm{c}) t=24.0 \mathrm{h}\).

Short answer

Answer: The ratio of decay rates for A and B is 4 at t = 0, t = 12 hours, and t = 24 hours.

Step-by-step solution

(Step 1: Recall the Decay formula )

The general decay formula is as follows $$ N(t) = N_0 e^{-\lambda t} $$ Where \(N(t)\) is the number of radioactive nuclei remaining after time \(t\), \(N_0\) is the initial number of radioactive nuclei, and \(\lambda\) is the decay constant.

(Step 2: Find Decay Constants for A and B)

We are given their half-lives, which we can use to find their decay constants by using the formula $$ \lambda = \frac{ln(2)}{t_{1/2}} $$ Thus, for A, the decay constant is $$ \lambda_A = \frac{ln(2)}{3} $$ And for B, the decay constant is $$ \lambda_B = \frac{ln(2)}{12} $$

(Step 3: Find the Decay Rates)

The decay rate or activity is given by the formula $$ R(t) = \lambda N(t) $$ For A, the decay rate at time \(t\) is $$ R_A(t) = \lambda_A N_A(t) = (\frac{ln(2)}{3}) N_{A0} e^{-\lambda_A t} $$ For B, the decay rate at time \(t\) is $$ R_B(t) = \lambda_B N_B(t) = (\frac{ln(2)}{12}) N_{B0} e^{-\lambda_B t} $$

(Step 4: Calculate the Ratio of Decay Rates for t=0)

For (a), we have \(t=0\), so our exponential function becomes 1. The ratio of the decay rates at \(t=0\) is: $$ \frac{R_A(0)}{R_B(0)} = \frac{\lambda_A N_{A0}}{\lambda_B N_{B0}} $$ Since A and B have equal initial numbers, \(N_{A0} = N_{B0}\). Therefore, $$ \frac{R_A(0)}{R_B(0)} = \frac{\lambda_A}{\lambda_B} = \frac{ln(2)/3}{ln(2)/12} = \frac{12}{3} = 4 $$

(Step 5: Calculate the Ratio of Decay Rates for t=12h and t=24h)

We now calculate the decay rate ratio for (b) \(t=12\,\text{h}\) and (c) \(t=24\,\text{h}\). $$ \frac{R_A(t)}{R_B(t)} = \frac{(\frac{ln(2)}{3}) N_{A0} e^{-\lambda_A t}}{(\frac{ln(2)}{12}) N_{B0} e^{-\lambda_B t}} = \frac{4 e^{ln(2)(\frac{1}{12} - \frac{1}{3})t}}{e^{ln(2)(\frac{1}{12} - \frac{1}{3})t}} = 4 $$ For both (b) and (c), since we have the same exponent in the numerator and denominator, they cancel out. Therefore, the ratio of decay rates for \(t=12\,\text{h}\) and \(t=24\,\text{h}\) is also 4. In conclusion, - (a) At \(t=0\), the ratio of decay rates \(R_A/R_B\) is 4. - (b) At \(t=12\,\text{h}\), the ratio of decay rates \(R_A/R_B\) is 4. - (c) At \(t=24\,\text{h}\), the ratio of decay rates \(R_A/R_B\) is 4.