Question

question 7 (mandatory) (1 point) a 90 g sample of plutonium - 238 has a half - life of 88 years. how long will it take for this sample to decay to 40 g?

Explanation:

Step1: Write the radioactive - decay formula

The radioactive - decay formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. Here, $N_0 = 90$ g, $N = 40$ g, and $T_{1/2}=88$ years.
So, $40 = 90(\frac{1}{2})^{\frac{t}{88}}$.

Step2: Rearrange the equation

First, divide both sides of the equation by 90: $\frac{40}{90}=(\frac{1}{2})^{\frac{t}{88}}$, which simplifies to $\frac{4}{9}=(\frac{1}{2})^{\frac{t}{88}}$.
Take the natural logarithm of both sides: $\ln(\frac{4}{9})=\ln((\frac{1}{2})^{\frac{t}{88}})$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{4}{9})=\frac{t}{88}\ln(\frac{1}{2})$.

Step3: Solve for $t$

We know that $\ln(\frac{4}{9})=\ln(4)-\ln(9)\approx1.386 - 2.197=- 0.811$ and $\ln(\frac{1}{2})=\ln(1)-\ln(2)=0 - 0.693=-0.693$.
Then, $t = 88\times\frac{\ln(\frac{4}{9})}{\ln(\frac{1}{2})}$.
Substitute the values of the logarithms: $t = 88\times\frac{-0.811}{-0.693}$.
$t = 88\times1.17$.
$t\approx103$ years.

Answer:

103 years