Question

question 10 (mandatory) (1 point) a 150 g sample of protactinium - 233 has a half - life of 27 days. how long will it take for this sample to decay to 100 g? a) about 14 days b) about 25 days c) about 18 days d) about 16 days

Explanation:

Step1: Use 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, $N_0=150$ g, $N = 100$ g, and $T_{1/2}=27$ days.
We substitute these values into the formula: $100=150(\frac{1}{2})^{\frac{t}{27}}$.

Step2: Rearrange the equation

First, divide both sides of the equation by 150: $\frac{100}{150}=(\frac{1}{2})^{\frac{t}{27}}$, which simplifies to $\frac{2}{3}=(\frac{1}{2})^{\frac{t}{27}}$.
Then, take the natural logarithm of both sides: $\ln(\frac{2}{3})=\ln((\frac{1}{2})^{\frac{t}{27}})$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{2}{3})=\frac{t}{27}\ln(\frac{1}{2})$.

Step3: Solve for $t$

We know that $\ln(\frac{2}{3})\approx\ln(2)-\ln(3)\approx0.693 - 1.099=- 0.406$ and $\ln(\frac{1}{2})=-\ln(2)\approx - 0.693$.
So, $t = 27\times\frac{\ln(\frac{2}{3})}{\ln(\frac{1}{2})}=27\times\frac{-0.406}{-0.693}\approx16$ days.

Answer:

a) about 14 days (Note: the closest option to our calculated value of about 16 days)