Question

calculate how long it would take for the following to decay to an activity of 1 becquerel (Bq).
1. A sample of cobalt-60(half-life - 5.27 years) whose original activity is 64 Bq.
2.A sample of iodine-131 (half-life - 8 days) whose original activity is 128 Bq.
3. A sample of polonium -210 (half-life - 138 days) whose original activity falls is 32 Bq.

Answer 1

Answer on Question 63530, Physics, Atomic and Nuclear Physics

Question:

Calculate how long it would take for the following to decay to an activity of 1 becquerel (Bq).

1) A sample of cobalt-60 (half-life - 5.27 years) whose original activity is 64 Bq.

2) A sample of iodine-131 (half-life - 8 days) whose original activity is 128 Bq.

3) A sample of polonium-210 (half-life - 138 days) whose original activity falls is 32 Bq.

Solution:

Let's use the famous formula for radioactive decay:

$$A = A _ {0} e ^ {- \lambda t},$$

here, $A_0$ is the original activity of the radioactive sample at time $t = 0$, $A$ is the activity of the radioactive sample at time $t$, $\lambda = \frac{0.693}{T_{1/2}}$ is the decay constant, $T_{1/2}$ is the half-life of the radioactive sample, $t$ is the time we are searching for.

From this formula we can find how long it would take for the following radioactive samples (cobalt-60, iodine-131, polonium-210) to decay to an activity of 1 Bq:

$$\begin{array}{l} \frac {A}{A _ {0}} = e ^ {- \lambda t}, \\ \ln \left(\frac {A}{A _ {0}}\right) = \ln \left(e ^ {- \lambda t}\right), \\ \ln \left(\frac {A}{A _ {0}}\right) = - \frac {0.693}{T _ {1 / 2}} t, \\ t = \left[ \frac {\ln \left(\frac {A}{A _ {0}}\right)}{(- 0.693)} \right] T _ {1 / 2}. \end{array}$$

1) For the sample of cobalt-60:

$$t = \left[ \frac {\ln \left(\frac {A}{A _ {0}}\right)}{(- 0 . 6 9 3)} \right] T _ {1 / 2} = \left[ \frac {\ln \left(\frac {1 B q}{6 4 B q}\right)}{(- 0 . 6 9 3)} \right] \cdot 5. 2 7 \text{ years} = 31. 6 \text{ years}.$$

2) For the sample of iodine-131:

$$t = \left[ \frac {\ln \left(\frac {A}{A _ {0}}\right)}{(- 0 . 6 9 3)} \right] T _ {1 / 2} = \left[ \frac {\ln \left(\frac {1 B q}{1 2 8 B q}\right)}{(- 0 . 6 9 3)} \right] \cdot 8 \text{ days} = 5 6 \text{ days}.$$

3) For the sample of polonium-210:

$$t = \left[ \frac {\ln \left(\frac {A}{A _ {0}}\right)}{(- 0 . 6 9 3)} \right] T _ {1 / 2} = \left[ \frac {\ln \left(\frac {1 B q}{3 2 B q}\right)}{(- 0 . 6 9 3)} \right] \cdot 1 3 8 \text{ days} = 6 9 0 \text{ days}.$$

Answer:

1) $t = 31.6$ years.

2) $t = 56$ days.

3) $t = 690$ days.

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