Question

At the beginning of an experiment, a scientist has 276 grams of radioactive goo. After 255 minutes, her sample has decayed to 17.25 grams.

Answer 1

ANSWER

$\begin{gathered} a)t1\text{/2}=63.75\min \\ b)G(t)=276\cdot(\frac{1}{2})^{\frac{t}{63.75}} \\ c)131.65g \end{gathered}$

EXPLANATION

a) At the start of the experiment, the scientist had 276 grams of radioactive goo and after 255 minutes, the sample decayed to 17.25 grams.

To find the half-life, we can apply the formula for half-life:

$N(t)=N_0(\frac{1}{2})^{\frac{t}{t1\text{/2}}_{}}$

where N0 = initial amount

N(t) = amount after time t

t = time elapsed

t1/2 =half-life

From the question, we have that:

$\begin{gathered} N(t)=17.25g \\ N_0=276g \\ t=255\text{ mins} \end{gathered}$

We have to find the half-life, t¹/₂. That is:

$\begin{gathered} 17.25=276(\frac{1}{2})^{\frac{255}{t1\text{/2}}} \\ \Rightarrow\frac{17.25}{276}=(\frac{1}{2})^{\frac{255}{t1\text{/2}}} \\ \Rightarrow0.0625=(\frac{1}{2})^{\frac{255}{t1\text{/2}}} \end{gathered}$

Converting to logarithmic function:

$\begin{gathered} \Rightarrow\log _{(\frac{1}{2})}0.0625=\frac{255}{t1\text{/2}} \\ \frac{255}{t1\text{/2}}=\frac{\log _{10}0.0625}{\log _{10}0.5} \\ \frac{255}{t1\text{/2}}=4 \\ \Rightarrow t1\text{/2=}\frac{255}{4} \\ t1\text{/2=63.75 mins} \end{gathered}$

Therefore, the half-life is 63.75mins.

b) Therefore, using the half-life, we have that the equation G(t) for the amount of goo remaining at time t is:

$G(t)=276\cdot(\frac{1}{2})^{\frac{t}{63.75}}$

c) To find the amount of goo remaining after 68 mins, we have to find G(t) when t = 68.

That is:

$\begin{gathered} G(t)=276\cdot(\frac{1}{2})^{\frac{68}{63.75}} \\ G(t)=276\cdot(\frac{1}{2})^{1.067} \\ G(t)=276\cdot0.477 \\ G(t)=131.65g \end{gathered}$

That is the amount remaining after 68 minutes.