Question

Element X decays radioactively with a half life of 14 minutes. If there are 160 grams

of Element X, how long, to the nearest tenth of a minute, would it take the element to
decay to 11 grams?
loudy
y = a(.5)^t/h

Answer 1

To find how long it would take for Element X to decay to 11 grams, we can use the radioactive decay equation y = a(0.5)^(t/h). Given that the half-life of Element X is 14 minutes, and the initial amount is 160 grams, we can substitute the values and solve for t when the remaining amount is 11 grams, which is approximately 43 minutes.

Step-by-step explanation:

To find how long it would take for Element X to decay to 11 grams, we can use the radioactive decay equation: y = a(0.5)^(t/h), where y is the remaining amount, a is the initial amount, t is the time, and h is the half-life.

Given that the half-life of Element X is 14 minutes, and the initial amount is 160 grams, we can substitute the values into the equation:

y = 160(0.5)^(t/14)

Now, we need to solve for t when y = 11:

11 = 160(0.5)^(t/14)

Divide both sides by 160:

0.06875 = (0.5)^(t/14)

Take the logarithm of both sides:

log(0.06875) = log((0.5)^(t/14))

Using logarithm properties, we can bring down the exponent:

log(0.06875) = (t/14) * log(0.5)

Solve for t:

t = (14 * log(0.06875)) / log(0.5)

Calculating this value gives us t ≈ 43 minutes (rounded to the nearest tenth of a minute).

Answer 2

`\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( (1)/(2) \right)^{(t)/(h)}\qquad \begin{cases} A=\textit{current amount}\dotfill &11\\ P=\textit{initial amount}\dotfill &160\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &14 \end{cases} \\\\\\ 11=160\left( (1)/(2) \right)^{(t)/(14)}\implies \cfrac{11}{160}=\left( \cfrac{1}{2} \right)^{(t)/(14)}\implies \log\left( \cfrac{11}{160} \right)=\log\left[ \left( \cfrac{1}{2} \right)^{(t)/(14)} \right]`

`\log\left( \cfrac{11}{160} \right)=\cfrac{t}{14}\log\left[ \left( \cfrac{1}{2} \right)\right]\implies \cfrac{\log\left( (11)/(160) \right)}{\log \left( (1)/(2) \right)}=\cfrac{t}{14} \\\\\\ 14\left( \cfrac{\log\left( (11)/(160) \right)}{\log \left( (1)/(2) \right)} \right)=t\implies {\LARGE \begin{array}{llll} \stackrel{mins}{54.1}\approx t \end{array}}`