Question #97b1f

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Question #97b1f

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Answer 1 · 2017-04-09T13:24:10

$= 13.879 g$ $= 13.879 " g"$

[to the nearest hundredth of a gram]

Explanation:

You'll need the Pu-239 half life which Google tells me is: $24, 100 years$ $24,100 " years"$

For exponential decay $A (t) = A_{o} e^{- k t}$ $A(t) = A_o e^(-kt)$ we can say this about the half life, $τ$ $tau$ :

$A (τ) = \frac{A_{o}}{2} = A_{o} e^{- k τ} \Rightarrow k = \frac{ln 2}{τ}$ $A(tau) = A_o/2 = A_o e^(- k tau ) implies k = (ln 2)/tau$

$A (t) = A_{o} e^{- \frac{ln 2}{τ} t}$ $A(t) = A_o e^(- (ln 2)/tau t)$

$A (t = 75, 000) = 120 \cdot e^{- \frac{ln 2}{24, 100} \cdot 75, 000}$ $A(t = 75,000) = 120 cdot e^(- (ln 2)/(24,100) cdot 75,000)$

$= 13.879 g$ $= 13.879 " g"$

Reality check, that is almost 3 half lives so we should get in the order of: $120 \cdot {(\frac{1}{2})}^{3} = 15 g$ $120 * (1/2)^(color(red)(3)) = 15 " g"$

Answer 2 · 2017-04-09T13:49:29

The mass is $= 13.891 g$ $=13.891g$

Explanation:

The half life of Plutonum 239 is $t_{\frac{1}{2}} = 2.4 \cdot 10^{4} y e a r s$ $t_(1/2)=2.4*10^4 years$

The radioactive decay constant is $λ = \frac{ln 2}{t_{\frac{1}{2}}}$ $lambda=ln2/(t_(1/2))$

So,

$λ = \frac{ln 2}{2.4 \cdot 10^{4}} = \frac{0.69}{2.4 \cdot 10^{4}}$ $lambda=ln2/(2.4*10^4 )=0.69/(2.4*10^4 )$

$= 0.2875 \cdot 10^{- 4} (y e a r s^{- 1})$ $=0.2875*10^-4 (years^-1)$

We apply the equation

$A = A_{0} \cdot e^{- λ t}$ $A=A_0*e^(-lamdat)$

The activity is proportional to the mass.

$m = m_{0} \cdot e^{- λ t}$ $m=m_0*e^(-lamdat)$

$m = 120 \cdot e^{- (0.2875 \cdot 10^{- 4} \cdot 75000)}$ $m=120*e^-(0.2875*10^-4*75000)$

$m = 120 \cdot e^{- 2.15625}$ $m=120*e^-2.15625$

$m = 120 \cdot 0.11576$ $m=120*0.11576$

$m = 13.891 g$ $m=13.891g$

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