Question

element x decays radioactively with a half - life of 11 minutes. if there are 370 grams of element x, how long, to the nearest tenth of a minute, would it take the element to decay to 4 grams? y = a(.5)^(t/h)

Explanation:

Step1: Identify decay formula

The radioactive - decay formula is $y = a(0.5)^{\frac{t}{h}}$, where $y$ is the final amount, $a$ is the initial amount, $t$ is the time elapsed, and $h$ is the half - life. Here, $a = 370$ grams, $y = 4$ grams, and $h = 11$ minutes.

Step2: Substitute values into formula

Substitute the values into the formula: $4=370(0.5)^{\frac{t}{11}}$.
First, divide both sides by 370: $\frac{4}{370}=(0.5)^{\frac{t}{11}}$. So, $0.01081=(0.5)^{\frac{t}{11}}$.

Step3: Take the natural logarithm of both sides

$\ln(0.01081)=\ln((0.5)^{\frac{t}{11}})$.
Using the property of logarithms $\ln(m^n)=n\ln(m)$, we get $\ln(0.01081)=\frac{t}{11}\ln(0.5)$.

Step4: Solve for $t$

We know that $\ln(0.01081)\approx - 4.52$ and $\ln(0.5)\approx - 0.693$.
Then, $t = 11\times\frac{\ln(0.01081)}{\ln(0.5)}$.
$t = 11\times\frac{- 4.52}{- 0.693}$.
$t = 11\times6.52$.
$t\approx71.7$ minutes.

Answer:

$71.7$ minutes