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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#