Question

The disintegration rate for a sample containing ${}_{27}^{60}Co$ as the only radioactive nuclide, is found to be $240$ atoms/ minute. $t_{1/2}$ of $Co$ is $5.2$ years. Find the number of atoms of $Co$ in the sample. How long must this radioactive sample be maintained before the rate falls to $100$ disintegration / minute.

Answer 1

${}_{27}^{60}Co$ has disintegration per min $=240$ atoms/minute

$t1/2=\left(Co\right)=5.2$ years

$A\left(dPS\right)=\frac{240}{60}=4$ atoms/ sec

$A=\lambda N$

$4=\frac{0.693}{5.2\times 365\times 24\times 3600}\times N$

$N=\frac{4\times 5.2\times 365\times 24\times 3600}{0.693}=9.465\times {10}^8$

$At=100dpm$

$\therefore At=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}\Rightarrow \frac{100}{240}={\left(\frac{1}{2}\right)}^{\frac{t}{5.2}}$

Solving by taking longarithm on both side, we have.

$t=6.6$ years