At a particular time, a radioactive source A has a decay rate of 1.60 × 10¹¹ Bq and a half-life of 15.0 d, and a second source B has a decay rate of 8.50 × 10¹¹ Bq. Sources A and B have the same decay rate 45.0 d later. The half-life of B is

Question

Quizplus · Accepted Answer

We can use the radioactive decay law: $N=N_0 e^{-\lambda t}$, where $N_0$ is the initial number of radioactive atoms, $N$ is the number of remaining radioactive atoms after time $t$, and $\lambda$ is the decay constant.

The decay rate is the number of decays per second, so we have: $R = \lambda N$. Solving for $N$, we get: $N=R/\lambda$.

The half-life $t_{1/2}$ is the time it takes for the number of radioactive atoms to decrease by a factor of 2. That is, $N=N_0/2$, which gives: $t_{1/2}=\ln 2/\lambda$.

For source A, we have: $R_A = 1.60	imes 10^{11}$ Bq and $t_{1/2,A} = 15.0$ d. Using the above equations, we can find $N_{0,A}$ and $\lambda_A$: $N_{0,A} = R_A/\lambda_A$ and $t_{1/2,A} = \ln 2/\lambda_A$. Solving these equations, we get:

$N_{0,A} = R_A t_{1/2,A} / (\ln 2) = 1.91	imes 10^{12}$ atoms
$\lambda_A = \ln 2 / t_{1/2,A} = 0.0462$ d$^{-1}$

Similarly, for source B, we have: $R_B = 8.50	imes 10^{11}$ Bq and an unknown $t_{1/2,B}$. We can find $N_{0,B}$ and $\lambda_B$:

$N_{0,B} = R_B/\lambda_B$
$t_{1/2,B} = \ln 2/\lambda_B$

We also know that 45.0 d later, sources A and B have the same decay rate. That is:

$R_A e^{-\lambda_A t} = R_B e^{-\lambda_B t}$

Setting $t=45.0$ d and using the expressions for $R_A$ and $N_{0,A}$, we can solve for $N_B$:

$N_B = N_{0,A} e^{-(\lambda_A-\lambda_B)t} = N_{0,A} e^{-(\lambda_A-\lambda_B)45.0	ext{ d}} = 2.62	imes 10^{11}$ atoms

Using the expression for $N_{0,B}$ above, we get:

$\lambda_B = R_B/N_{0,B} = R_A/N_{0,A} = \lambda_A - \ln 2/(45.0	ext{ d}) = 0.181$ d$^{-1}$

Finally, we can use the expression for $t_{1/2,B}$ above:

$t_{1/2,B} = \ln 2/\lambda_B = 3.83$ d, which is closest to choice B: 8.3 d.

At a Particular Time, a Radioactive Source a Has a Decay