Question

lendium-176 has a half-life of 3.85 × 10¹⁰ years. after 1.155 × 10¹¹ years, how much lendium-176 will remain from an original 16.8 g sample?
○ 2.10 g
○ 3.00 g
○ 5.56 g
○ 8.40 g

Explanation:

Step 1: Recall the half - life formula

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

Given that $N_0 = 16.8\space g$, $T_{1/2}=3.85\times 10^{10}\space years$, and $t = 1.155\times 10^{11}\space years$.

Step 2: Calculate the number of half - lives ($n$)

First, we calculate the number of half - lives $n=\frac{t}{T_{1/2}}$. Substitute the given values:
\[

$$\begin{align*} n&=\frac{1.155\times 10^{11}}{3.85\times 10^{10}}\\ &=\frac{1.155}{3.85}\times10^{11 - 10}\\ & = 0.3\times10^{1}\\ &=3 \end{align*}$$

\]

Step 3: Calculate the remaining amount ($N$)

Now, use the formula $N = N_0\times(\frac{1}{2})^n$. Substitute $N_0 = 16.8\space g$ and $n = 3$:
\[

$$\begin{align*} N&=16.8\times(\frac{1}{2})^3\\ &=16.8\times\frac{1}{8}\\ & = 2.10\space g \end{align*}$$

\]

Answer:

2.10 g (corresponding to the option "2.10 g")