Question

6 radioactive decay

1. thorium - 232 has a half - life of about 14 billion years. if you had 1 g of th - 232, how long would you have to wait until it had decayed to 0.25 g?
2. carbon 14 has a half - life of 5730 years. what is its decay constant?
3. if you found a fossilized plant that contained 10% of the carbon - 14 found in the living version of the plant, how old is the fossil? (use your answer from problem 2 to find the solution).
4. bananas contain potassium - 40, which is radioactive with a half - life of 40.3×10^15 s. what is the decay constant for potassium - 40? if there were 2×10^22 potassium - 40 atoms in your banana, how many would decay over the course of one week?

Explanation:

1.

Step1: Recall half - life formula

The decay formula is $N = N_0e^{-\lambda t}$, and at half - life $t_{1/2}$, $N=\frac{N_0}{2}$. So, $\frac{1}{2}=e^{-\lambda t_{1/2}}$. Taking the natural logarithm of both sides, we get $\ln(\frac{1}{2})=-\lambda t_{1/2}$, and $\lambda=\frac{\ln 2}{t_{1/2}}$. Given $t_{1/2}=14\times10^{9}$ years.

Step2: Calculate decay constant

$\lambda=\frac{\ln 2}{14\times10^{9}\text{ years}}\approx\frac{0.693}{14\times10^{9}\text{ years}}\approx 4.95\times 10^{-11}\text{ year}^{-1}$.

Step3: Use decay formula to find time

We want to find $t$ when $N = 0.25N_0$. Substitute into $N = N_0e^{-\lambda t}$, we get $0.25N_0=N_0e^{-\lambda t}$, or $0.25 = e^{-\lambda t}$. Taking the natural logarithm of both sides: $\ln(0.25)=-\lambda t$. Since $\ln(0.25)=\ln(\frac{1}{4})=- \ln 4=-2\ln 2$. Then $t=\frac{2\ln 2}{\lambda}$. Substituting $\lambda=\frac{\ln 2}{14\times10^{9}\text{ years}}$, we get $t = 28\times10^{9}$ years.

Step1: Recall decay constant formula

Using $\lambda=\frac{\ln 2}{t_{1/2}}$, with $t_{1/2}=5730$ years.

Step2: Calculate decay constant

$\lambda=\frac{\ln 2}{5730\text{ years}}\approx\frac{0.693}{5730\text{ years}}\approx1.21\times 10^{-4}\text{ year}^{-1}$

Step1: Use decay formula for carbon - 14

The decay formula is $N = N_0e^{-\lambda t}$, where $\lambda=\frac{\ln 2}{t_{1/2}}$ and $t_{1/2} = 5730$ years. We know that $N = 0.1N_0$.

Step2: Substitute into decay formula

$0.1N_0=N_0e^{-\lambda t}$, so $0.1=e^{-\lambda t}$. Taking the natural logarithm of both sides: $\ln(0.1)=-\lambda t$. Since $\lambda=\frac{\ln 2}{5730\text{ years}}$, then $t=-\frac{\ln(0.1)}{\lambda}=-\frac{\ln(0.1)}{\frac{\ln 2}{5730\text{ years}}}$.

Step3: Calculate the age

$t=\frac{\ln(10)}{\frac{\ln 2}{5730\text{ years}}}\approx\frac{2.303}{\frac{0.693}{5730\text{ years}}}\approx18995$ years

Answer:

$28$ billion years

2.