Suggested languages for you:

Americas

Europe

Q44P

Expert-verified
Found in: Page 1304

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 42-19 shows the decay of parents in a radioactive sample. The axes are scaled by ${{\mathbit{N}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbit{x}}{{\mathbf{10}}}^{{\mathbf{6}}}$ and ${{\mathbit{t}}}_{{\mathbf{S}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbit{s}}$. What is the activity of the sample at?

The activity of the sample at t = 27s is 60 Bq.

See the step by step solution

## Step 2: Determine the formula for the decay

Write the formula for undecayed sample remaining after a given time, as follows:

$N={N}_{0}{e}^{-\lambda t}$ …… (i)

Write the formula for the rate of decay:

$R=\lambda N$

Here, $\lambda$ is the disintegration constant, N is the number of undecayed nuclei.

## Step 3: Calculate the activity of the sample at 27h

The number of atoms present initially at t = 0 is ${\mathrm{N}}_{0}=2×{10}^{6}$. From Fig. 42-19, determine that the number is halved at t = 2s. Thus, using equation (i), determine the disintegration constant of the sample as follows:

Now, using this value in equation (i), determine value of the remaining nuclei of the radioactive sample at $t=27s$ as follows:

$N=\left(2×{10}^{6}\right){e}^{-\left(0.3466{s}^{-1}\right)\left(27s\right)}\phantom{\rule{0ex}{0ex}}=173$

Now, the activity of the sample at the given time is given using the above values in equation (ii) as follows:

$R=\left(0.3466{s}^{-1}\right)\left(173\right)\phantom{\rule{0ex}{0ex}}\approx 60{s}^{-1}\phantom{\rule{0ex}{0ex}}=60Bq$

Hence, the activity of the sample is 60 Bq.