How much of a sample of cesium-137 \(\left(t_{1 / 2}=30\right.\) years) must have been present originally if, after 270 years, \(15.0 \mathrm{~g}\) remain?

Understanding radioactive decay is essential to grasp how certain materials change over time. Radioactive decay occurs when an unstable atomic nucleus loses energy by emitting radiation.
This process can be spontaneous and results in the formation of stable or sometimes different unstable elements.
It plays a critical role in many fields such as geology (dating rocks), medicine (treating cancer), and archaeology (dating artifacts).
The rate at which a radioactive substance decays is mathematically predictable using the concept of half-life.

The half-life formula helps us determine how long it takes for half of a radioactive substance to decay.
The formula is expressed as:
\[ N = N_0 \times \frac{1}{2}^{\frac{t}{t_{1/2}}} \] Here’s a breakdown of what each symbol represents:

• \( N \): Amount of the substance left after time \( t \)
• \( N_0 \): Original amount of the substance
• \( t \): Elapsed time
• \( t_{1/2} \): Half-life of the substance

For instance, Cesium-137 has a half-life (\( t_{1/2} \)) of 30 years.
This means in 30 years, only half of the Cesium-137 remains un-decayed.By plugging given values into the formula, we can effortlessly predict the remaining quantity after a certain period.

In problems involving radioactive decay, we sometimes need to work backwards to find the original quantity of a substance.
Let's take the given exercise:

• After 270 years, we find that 15.0 g of Cesium-137 remains.
• The half-life \( t_{1/2} \) of Cesium-137 is 30 years.

Using the half-life formula, we start by plugging in the given values:
\[ 15.0 = N_0 \times \frac{1}{2}^{\frac{270}{30}} \]Next, we simplify the exponent: \[ \frac{270}{30} = 9 \]This converts our equation to: \[ 15.0 = N_0 \times \frac{1}{2}^9 \]Then, we calculate \( \frac{1}{2}^9 \): \[ \frac{1}{2}^9 = \frac{1}{512} \]So, the equation now is: \[ 15.0 = N_0 \times \frac{1}{512} \] Finally, we multiply both sides by 512 to solve for \( N_0 \): \[ 15.0 \times 512 = N_0 \]\[ N_0 = 7680 \,g \]Thus, the original quantity of Cesium-137 was \( 7680 \, g \). By understanding each part of the process, you can confidently solve similar problems.