Question

question 12 of 24 step 2 of 3 the half-life of gold-194 is approximately 1.6 days. step 2 of 3: how much of a 9 gram sample of gold-194 would remain after 2 days? round to three decimal places. answerhow to enter your answer (opens in new window) 2 points 01:57 keyboard)

Explanation:

Step 1: Recall the half - life formula

The formula for radioactive decay is $N(t)=N_0\times(\frac{1}{2})^{\frac{t}{T}}$, where $N(t)$ is the amount of the substance remaining after time $t$, $N_0$ is the initial amount of the substance, $T$ is the half - life of the substance.

Here, $N_0 = 9$ grams, $t = 2$ days, and $T=1.6$ days.

Step 2: Substitute the values into the formula

Substitute $N_0 = 9$, $t = 2$, and $T = 1.6$ into the formula:

$N(2)=9\times(\frac{1}{2})^{\frac{2}{1.6}}$

First, calculate the exponent $\frac{2}{1.6}=\frac{20}{16}=\frac{5}{4} = 1.25$

Then, $(\frac{1}{2})^{1.25}=2^{- 1.25}$. We know that $a^{-n}=\frac{1}{a^{n}}$ and $a^{m + n}=a^{m}\times a^{n}$. Also, $2^{1.25}=2^{1+\frac{1}{4}}=2^{1}\times2^{\frac{1}{4}}=2\times\sqrt[4]{2}\approx2\times1.1892 = 2.3784$

So, $2^{-1.25}=\frac{1}{2.3784}\approx0.4205$

Now, $N(2)=9\times0.4205 = 3.7845\approx3.785$ (rounded to three decimal places)

Answer:

$3.785$