The half-life of a radioactive substance is 97 years. If 10 mg are produced initially, we need to find how long it takes for only 1 mg to remain.

1. Problem Description

The half-life of a radioactive substance is 97 years. If 10 mg are produced initially, we need to find how long it takes for only 1 mg to remain.

2. Solution Steps

The formula for radioactive decay is:

$N(t) = N_0 * (1/2)^{(t/T)}$

where:

$N(t)$ is the amount of substance remaining after time $t$.

$N_0$ is the initial amount of substance.

$t$ is the time elapsed.

$T$ is the half-life of the substance.

In this case:

$N(t) = 1$ mg

$N_0 = 10$ mg

$T = 97$ years

We want to find $t$.

$1 = 10 * (1/2)^{(t/97)}$

Divide both sides by 10:

$1/10 = (1/2)^{(t/97)}$

Take the logarithm of both sides (base 1/2 or base 10 or natural logarithm):

$log(1/10) = (t/97) * log(1/2)$

$log(1/10) = log(10^{-1}) = -1$

$log(1/2) = log(2^{-1}) = -log(2)$

$-1 = (t/97) * (-log(2))$

$1 = (t/97) * log(2)$

$t/97 = 1/log(2)$

$t = 97 / log(2)$

Using base 10 logarithm:

$log_{10}(2) \approx 0.3010$

$t = 97 / 0.3010 \approx 322.25$

Using natural logarithm:

$ln(2) \approx 0.6931$

$t = 97 / 0.6931 \approx 140$

Then we can also think about it this way:

$10 \xrightarrow{97 \text{ years}} 5 \xrightarrow{97 \text{ years}} 2.5 \xrightarrow{97 \text{ years}} 1.25 \xrightarrow{97 \text{ years}} 0.625$

Since 1 is between 1.25 and 0.625, it is between 3 half-lives and 4 half-lives.

After approximately $t = 3*97 = 291$ years, we expect the substance remaining to be $10*(1/2)^3 = 10/8 = 1.25$ mg

After approximately $t = 4*97 = 388$ years, we expect the substance remaining to be $10*(1/2)^4 = 10/16 = 0.625$ mg

We need to find the number of half-lives required.

$(1/2)^n = 1/10$

$2^n = 10$

$n = log_2(10) = log(10)/log(2) \approx 1/0.301 = 3.32$

Time = $3.32 * 97 = 322.04$

3. Final Answer

322 years