Answer
${\bf 86.14}$
Work Step by Step
- The half-life of $ ^{235}\text{U} $ is 700 million years.
- The age of the Earth is approximately 4.5 billion years.
We need to calculate the "then-to-now" ratio of $ ^{235}\text{U} $, representing how many times more $ ^{235}\text{U} $ atoms were present when the Earth formed compared to now.
The number of $ ^{235}\text{U} $ atoms remaining after time $ t $ is given by
$$
N_U = (N_U)_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
where:
- $ N_U $ is the current number of $ ^{235}\text{U} $ atoms.
- $ (N_U)_0 $ is the number of $ ^{235}\text{U} $ atoms when the Earth formed.
- $ t $ is the time elapsed (4.5 billion years).
- $ t_{1/2} $ is the half-life of $ ^{235}\text{U} $ (700 million years).
The ratio of $ ^{235}\text{U} $ atoms when the Earth formed to the number now is given by:
$$
\frac{(N_U)_0}{N_U} = \frac{(N_U)_0 }{(N_U)_0\left( \frac{1}{2} \right)^{t / t_{1/2}}}= \frac{1}{\left( \frac{1}{2} \right)^{t / t_{1/2}}}
$$
Thus,
$$
\frac{(N_U)_0}{N_U} = \left( 2 \right)^{t / t_{1/2}}
$$
where $ t = 4.5 $ billion years =$4500$ million years and $ t_{1/2} = 700 $ million years:
$$
\frac{(N_U)_0}{N_U} = \left( 2 \right)^{4500/700 }=\color{red}{\bf 86.14}
$$
This means that the abundance of $ ^{235}\text{U} $ was about 86 times greater when the Earth formed compared to today.
