The decay rate of radio active material at any time \(t\) is proportional to its mass at that time. The mass is 27 grams when \(t=0\). After three hours it was found that 8 grams are left. Then the substance left after one more hour is

Let ' \(x\) ' be the mass of the material at time ' \(t\) '.
\(\therefore \frac{\mathrm{d} x}{\mathrm{dt}}=-\mathrm{k} x,(-\mathrm{ve}\) sign indicates decay. \()\)
\(\therefore \int \frac{\mathrm{d} x}{x}=-\mathrm{k} \int \mathrm{dt}\)
\(\therefore \log |x|=-\mathrm{kt}+\mathrm{c}\)
When \(\mathrm{t}=0, x=27\)
\(\therefore \mathrm{c}=\log 27\)
\(\therefore \log |x|=-\mathrm{kt}+\log 27\)
When \(\mathrm{t}=3, x=8\)
\(
\therefore \mathrm{k}=\log \left(\frac{3}{2}\right)
\)
When \(\mathrm{t}=4\), we get
\(\log |x|=-4 \log \left(\frac{3}{2}\right)+\log 27 \)
\( \therefore \log |x|=\log \left(\frac{16}{3}\right) \)
\( \therefore x=\frac{16}{3} \text { grams }\)