# Question #97b1f

Apr 9, 2017

$= 13.879 \text{ g}$

[to the nearest hundredth of a gram]

#### Explanation:

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

For exponential decay $A \left(t\right) = {A}_{o} {e}^{- k t}$ we can say this about the half life, $\tau$:

$A \left(\tau\right) = {A}_{o} / 2 = {A}_{o} {e}^{- k \tau} \implies k = \frac{\ln 2}{\tau}$

$A \left(t\right) = {A}_{o} {e}^{- \frac{\ln 2}{\tau} t}$

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

$= 13.879 \text{ g}$

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

Apr 9, 2017

The mass is $= 13.891 g$

#### Explanation:

The half life of Plutonum 239 is ${t}_{\frac{1}{2}} = 2.4 \cdot {10}^{4} y e a r s$

The radioactive decay constant is $\lambda = \ln \frac{2}{{t}_{\frac{1}{2}}}$

So,

$\lambda = \ln \frac{2}{2.4 \cdot {10}^{4}} = \frac{0.69}{2.4 \cdot {10}^{4}}$

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

We apply the equation

$A = {A}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

The activity is proportional to the mass.

$m = {m}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

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

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

$m = 120 \cdot 0.11576$

$m = 13.891 g$