Question

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioactive nuclide is left after 10 half-lives? How long will it take for 10 half-lives to pass for plutonium-239?

Answer 1

Answer:

• 0.09 % of the original radioactive nucllde its left after 10 half-lives
• It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}}$,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }$

$N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Answer 2

Answer:

percentage left = 10 %

241 000 years

Explanation:

The half life, $t_{\frac{1}{2} }$ of plutonium-239 is 24 100 years.

Thus, the number of half lives that result in the decrease of the nucleotide is given by the following equation:

The percentage left will be 10 %

if $t_{\frac{1}{2} } = 24 100$, then after 10 half-life, the decay will be:

$t_{10 years} = 24 100* 10\\ = 241 000$