Question

Suppose that $20$ days is half-life of a particular substance and there are only $10$ grams of the substance available. How much of the approximate substance remains after $75$ days?

• A. $0.743$ grams
• B. $0.591$ grams
• C. $0.523$ grams
• D. $0.436$ grams

Answer 1

The correct option is A $0.743$ grams

Decay rate of the substance is given by,

$\frac{dx}{dt}=-kx$ where $x=$ mass of substance remaining

$\Rightarrow \frac{dx}{x}=-kdt$

$\Rightarrow \underset{x_0}{\overset{x}{\int }}\frac{dx}{x}=-\underset{0}{\overset{t}{\int }}kdt$

$\Rightarrow {\left[\ln x\right]}_{x_0}^x=-kt$

$\Rightarrow \ln x-\ln x_0=-kt$

$\Rightarrow \ln \left(\frac{x}{x_0}\right)=-kt$---------------$\left(1\right)$

For half life, $x=\frac{x_0}{2}$,

So, $\ln \left(\frac{x_0/2}{x_0}\right)=-k{t}_{1/2}$

$\Rightarrow \ln \left(\frac{1}{2}\right)=-k{t}_{1/2}$

$\Rightarrow k=\frac{-\ln 2}{-t_{1/2}}$

$=\frac{\ln 2}{20}$

Now, we have to find $x$ after $t=75days$.

$x_0=10g$

Using equation (1),

$\ln \left(\frac{x}{10}\right)=-\left(\frac{\ln 2}{20}\right)\times 75$

$\Rightarrow x=10e^{(-\frac{15\ln 2}{4})}$

$=10\times 2^{\left(\frac{-15}{4}\right)}$

$=0.743grams$