Question

element x decays radioactively with a half - life of 7 minutes. if there are 150 grams of element x, how long, to the nearest tenth of a minute, would it take the element to decay to 3 grams?

Explanation:

Step1: Recall 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.
We know that $N_0 = 150$ grams, $N = 3$ grams, and $T_{1/2}=7$ minutes.

Step2: Substitute values into the formula

Substitute the values into the formula: $3 = 150(\frac{1}{2})^{\frac{t}{7}}$.
First, divide both sides by 150: $\frac{3}{150}=(\frac{1}{2})^{\frac{t}{7}}$.
Simplify $\frac{3}{150}=\frac{1}{50}$, so $\frac{1}{50}=(\frac{1}{2})^{\frac{t}{7}}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{1}{50})=\ln((\frac{1}{2})^{\frac{t}{7}})$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{1}{50})=\frac{t}{7}\ln(\frac{1}{2})$.
We know that $\ln(\frac{1}{50})=-\ln(50)\approx - 3.912$ and $\ln(\frac{1}{2})=-\ln(2)\approx - 0.693$.

Step4: Solve for $t$

$-3.912=\frac{t}{7}\times(- 0.693)$.
Multiply both sides by 7: $-3.912\times7=-0.693t$.
$-27.384=-0.693t$.
Divide both sides by $-0.693$: $t=\frac{27.384}{0.693}\approx39.5$.

Answer:

$39.5$ minutes