Strontium-90 has a half-life of 28 years. If a sample was tested in 1980 and found to be emitting 240 counts/min, in what year would the same sample be found to be emitting 30 counts/min? How much of the original Sr-90 would be left?

The concept of half-life is crucial in understanding radioactive decay problems. Half-life is defined as the period it takes for half of the radioactive atoms in a sample to decay and transform into another element or isotope. For instance, if you start with 100 units of a radioactive substance, after one half-life, only 50 units will remain. After another half-life, 25 units will remain, and so on.

For this specific exercise, Strontium-90 (Sr-90) has a half-life of 28 years. This means that every 28 years, the amount of Sr-90 in the sample reduces to half of its previous amount.

• Initial amounts: Let's say you begin with 100 units of Sr-90 in 1980.
• First half-life: By 2008 (28 years later), you will have 50 units remaining.
• Second half-life: By 2036, only 25 units will remain.
• Third half-life: By 2064, only 12.5 units will remain, which is exactly 12.5% of the original quantity.

This reduction over time helps scientists predict the behavior of radioactive materials.

The radioactive decay formula is a mathematical representation of how much of a radioactive substance remains after a certain period of time. The formula is:

\[ N(t) = N_0 \times \frac{1}{2}^{(t/T)} \]

Where:

• \(N(t)\) is the amount remaining after time \(t\)


• \(N_0\) is the initial amount


• \(T\) is the half-life of the substance

To solve problems like the one in the exercise, it's essential to apply this formula accurately. For example, in the case of Sr-90:

In 1980, the count rate was 240 counts/min. We need to find when the count rate will decrease to 30 counts/min. Plugging the values into the formula:

\[ 30 = 240 \times \frac{1}{2}^{(t/28)} \]
\[ \frac{30}{240} = \frac{1}{2}^{(t/28)} \]
\[ \frac{1}{8} = \frac{1}{2}^{(t/28)} \]
By recognizing that \(\frac{1}{8} = \frac{1}{2}^3\), we set the exponents equal to solve for \(t\):
\[ 3 = \frac{t}{28} \]
\[ t = 84 \text{ years} \]
Thus, the sample will decay to 30 counts/min in 84 years, which added to 1980 gives us the year 2064.

Strontium-90 (Sr-90) is a radioactive isotope that is produced during nuclear fission of uranium and plutonium in nuclear reactors and during nuclear weapons explosions. It has a half-life of 28 years and decays by beta emission, transforming into yttrium-90.

• Health Impacts: Sr-90 can pose serious health risks. It behaves like calcium in the body and can be deposited in bones and teeth if ingested. Prolonged exposure can increase the risk of bone cancer and leukemia.


• Environmental Concerns: Due to its long half-life, Sr-90 can persist in the environment for a significant time, making it a long-term threat when released in large quantities, like in the Chernobyl disaster.

In the context of the exercise, understanding Sr-90's decay helps to calculate how much of the isotope remains over decades. As calculated, after 84 years, which equates to 3 half-lives, only 12.5% of the original sample remains.

This highlights the persistent nature of Sr-90 in the environment and the importance of monitoring its levels in areas affected by nuclear activities.