Question

5. The activity of certain radio nuclide decrease to 15% of its original value in 10 days. fine its half life?

Answer 1

Given:

Decrease in a radioactive nuclide = 15%

Time for the radioactive decay = 10 days

To find:

Half life of the radioactive element

Solution:

Let the amount of a radioactive element = x

Amount after the decay = 0.15x

Duration for the decay = 10 days

Half life period of the element = $t_{\frac{1}{2}}$

Formula for the Final amount after decay is,

$N(t)=N_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

0.15x = $x(\frac{1}{2})^{(\frac{10}{t_{\frac{1}{2}}})}$

0.15 = $(\frac{1}{2})^{(\frac{10}{t_{\frac{1}{2}}})}$

log(0.15) = $\text{log}[(\frac{1}{2})^{(\frac{10}{t_{\frac{1}{2}}})}]$

-0.8239 = $\text{log}[(2)^{-(\frac{10}{t_{\frac{1}{2}}})}]$

-0.8239 = $-\frac{10}{t_{\frac{1}{2}}}(log2)$

0.8239 = $\frac{3.0103}{t_{\frac{1}{2}}}$

$t_{\frac{1}{2}}$ = $\frac{3.0103}{0.8239}$

   = 3.65 days

Therefore, half life of the radioactive element is 3.65 days.