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Answer 1 · 2017-04-09T13:24:10

$= 13.879 " g"$

[to the nearest hundredth of a gram]

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

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

$A(tau) = A_o/2 = A_o e^(- k tau ) implies k = (ln 2)/tau$

$A(t) = A_o e^(- (ln 2)/tau t)$

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

$= 13.879 " g"$

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

Answer 2 · 2017-04-09T13:49:29

The mass is $=13.891g$

The half life of Plutonum 239 is $t_(1/2)=2.4*10^4 years$

The radioactive decay constant is $lambda=ln2/(t_(1/2))$

So,

$lambda=ln2/(2.4*10^4 )=0.69/(2.4*10^4 )$

$=0.2875*10^-4 (years^-1)$

We apply the equation

$A=A_0*e^(-lamdat)$

The activity is proportional to the mass.

$m=m_0*e^(-lamdat)$

$m=120*e^-(0.2875*10^-4*75000)$

$m=120*e^-2.15625$

$m=120*0.11576$

$m=13.891g$