To determine how long it takes for \( \frac{15}{16} \) of a sample of Lead-202 to decay, we first need to understand what this means in terms of the remaining fraction of the sample.
If \( \frac{15}{16} \) of the sample has decayed, then \( \frac{1}{16} \) remains.
The decay of a radioactive substance can be described using the half-life formula. The remaining quantity of a radioactive substance after a certain amount of time can be described by the formula:
\[ N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life.
In this case, we want \( N(t) = \frac{1}{16} N_0 \) and \( T_{1/2} = 53,000 \) years.
Setting up the equation:
\[ \frac{1}{16} N_0 = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{53,000}} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{t}{53,000}} \]
Now, we can express \( \frac{1}{16} \) in terms of powers of \( \frac{1}{2} \):
\[ \frac{1}{16} = \left( \frac{1}{2} \right)^4 \]
This gives us:
\[ \left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{\frac{t}{53,000}} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 4 = \frac{t}{53,000} \]
Now we can solve for \( t \):
\[ t = 4 \times 53,000 = 212,000 \text{ years} \]
Therefore, the time it takes for \( \frac{15}{16} \) of a sample of Lead-202 to decay is 212,000 years.