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Q 3.2-23E

Expert-verifiedFound in: Page 101

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In Problems 23–27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive substance is the time it takes for one-half of the substance to disintegrate. If initially there are 50 g of a radioactive substance and after 3 days there are only 10 g remaining, what percentage of the original amount remains after 4 days?**

** **

After 4 days, the remaining radioactive substance will be **11.7%** of the original amount.

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present. Let the present amount of the radioactive substance be N.

Therefore, $\frac{dN}{dt}\propto N$

Given that there are 50g of a radioactive substance and after 3 days there are only 10g remain. We have to find the mass of the substance, remaining after 4 days, and its percentage of the original amount.

Given,

$\frac{dN}{dt}\propto N\phantom{\rule{0ex}{0ex}}\frac{dN}{dt}=-\lambda N\phantom{\rule{0ex}{0ex}}$

where, $\lambda $is the constant of proportionality.

$\frac{dN}{N}=-\lambda \phantom{\rule{0ex}{0ex}}\int \frac{dN}{N}=-\lambda \int dt\phantom{\rule{0ex}{0ex}}lnN=-\lambda t+ln{N}_{0}\phantom{\rule{0ex}{0ex}}$

** **where, In N_{0} is an arbitrary constant.

$lnN-ln{N}_{0}=-\lambda t\phantom{\rule{0ex}{0ex}}ln\left(\frac{N}{{N}_{0}}\right)=-\lambda t\phantom{\rule{0ex}{0ex}}\frac{N}{{N}_{0}}={e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}N={N}_{0}{e}^{-\lambda t}\xb7\xb7\xb7\xb7\xb7\xb7\left(1\right)\phantom{\rule{0ex}{0ex}}$

One will use this formula to solve the question.

Let the initial amount of the radioactive substance be i.e., N_{0}= 50g and given that the remaining amount of radioactive substance after 3 days is 10g i.e.,

t = 3 days and N = 10g

Now, from the equation (1),

$10=50{e}^{-3\lambda}\phantom{\rule{0ex}{0ex}}\frac{1}{5}={e}^{-3\lambda}\phantom{\rule{0ex}{0ex}}{e}^{3\lambda}=5\phantom{\rule{0ex}{0ex}}3\lambda =ln5\phantom{\rule{0ex}{0ex}}\lambda =\frac{ln5}{3}\phantom{\rule{0ex}{0ex}}\lambda =0.5365\phantom{\rule{0ex}{0ex}}$

One will use this value of $\lambda $ in the next step to find the value of the mass of the remaining radioactive substance after 4days.

Now we will find the mass of the remaining radioactive substance after 4 days.

For this, let N be the mass to be found,

N_{0}=50 g

Time, t = 4 days

(From Step 3)

Using the equation (1),

$N={N}_{0}{e}^{-\lambda t}\phantom{\rule{0ex}{0ex}}N=\left(50\right)\xb7{e}^{-\left(0.5365\right)4}\phantom{\rule{0ex}{0ex}}N=5.847g\phantom{\rule{0ex}{0ex}}$

**Hence, the mass of the remaining radioactive substance after 4 days is 5.847 g**.

The mass of the remaining radioactive substance after 4 days is 5.847 g

Therefore,

Percentage of remaining mass

$=\frac{5.847}{{N}_{0}}\times 100\phantom{\rule{0ex}{0ex}}=\frac{5.847}{50}\times 100\phantom{\rule{0ex}{0ex}}=11.7\%\phantom{\rule{0ex}{0ex}}$

**Thus, the percentage of the original amount that remains after 4 days is 11.7%.**

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