Physics, 02.09.2019 16:30 diontna29

Apractical 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?

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Answer from: zay65307

0.09 % of the original radioactive nucllde its left after 10 half-livesIt 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 from: 21gl13826

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:

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

Answer from: Quest

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.

Answer from: Quest

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic = - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

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