Suggested languages for you:

Americas

Europe

Q44P

Expert-verifiedFound in: Page 1304

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 ${{\mathit{N}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathit{x}}{{\mathbf{10}}}^{{\mathbf{6}}}$**** and ${{\mathit{t}}}_{{\mathbf{S}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathit{s}}$****. What is the activity of the sample at****?**

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

**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.

The number of atoms present initially at t = 0 is ${\mathrm{N}}_{0}=2\times {10}^{6}$. 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:

$\lambda =\frac{1}{t}\mathrm{ln}\left(\frac{{N}_{0}}{N}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2s}\mathrm{ln}\left(\frac{{N}_{0}}{\frac{{N}_{0}}{2}}\right)\phantom{\rule{0ex}{0ex}}=0.3466{s}^{-1}$

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

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

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.

94% of StudySmarter users get better grades.

Sign up for free