To solve this problem, we can use the concept of exponential decay. The mass of the radioactive isotope decreases by half every 18 years, so we can express the decay using the following formula:

$M = M_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}$

Where:

• $M$ is the final mass,
• $M_0$ is the initial mass,
• $t$ is the time in years,
• $T$ is the half-life period.

Given:

• $M = 210$ grams,
• $M_0 = 500$ grams,
• $T = 18$ years.

We need to find $t$, the time required for the mass to decay to 210 grams.

Substitute the given values into the formula:

$210 = 500 \times \left(\frac{1}{2}\right)^{\frac{t}{18}}$

First, divide both sides by 500:

$\frac{210}{500} = \left(\frac{1}{2}\right)^{\frac{t}{18}}$

Simplify the left side:

$0.42 = \left(\frac{1}{2}\right)^{\frac{t}{18}}$

To solve for $t$, we take the natural logarithm (ln) of both sides:

$\ln(0.42) = \frac{t}{18} \times \ln\left(\frac{1}{2}\right)$

Now, solve for $t$:

$t = 18 \times \frac{\ln(0.42)}{\ln\left(\frac{1}{2}\right)}$

Let's calculate that:

$t = 18 \times \frac{\ln(0.42)}{\ln(0.5)}$

Using the values $\ln(0.42) \approx -0.8675$ and $\ln(0.5) \approx -0.6931$:

$t = 18 \times \frac{-0.8675}{-0.6931} \approx 18 \times 1.2515 \approx 22.5 \text{ years}$

So, it will take approximately 22.5 years for the mass of the sample to reach 210 grams.

Would you like more details on this, or do you have any questions?

Here are some related questions:

1. How long would it take for the mass to decay to 100 grams?
2. What would the mass be after 36 years?
3. How does changing the half-life affect the decay time?
4. What if the initial mass was 1000 grams?
5. How can this formula be applied to other types of decay problems?

Tip: When dealing with half-life problems, using logarithms helps solve for time, especially when the decay doesn't result in an exact half-life period.