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The half-lives of different radioisotopes are given in the table below.

\begin{tabular}{|c|c|}
\hline Radioisotope & Half-life (min) \\
\hline argon-44 & 12 \\
\hline lead-196 & 37 \\
\hline potassium-44 & 22 \\
\hline indium-117 & 43 \\
\hline
\end{tabular}

If the initial amount of potassium-44 is 7.2 g, how much potassium-44 is left in the body after 44 min?

Mass: [tex]$\square$[/tex] g


Sagot :

To determine how much potassium-44 remains in the body after 44 minutes, we can use the concept of half-life. Here is a step-by-step solution:

1. Identify the given data:
- Initial mass of potassium-44 ([tex]\( m_0 \)[/tex]): 7.2 grams
- Half-life of potassium-44 ([tex]\( t_{1/2} \)[/tex]): 22 minutes
- Time elapsed ([tex]\( t \)[/tex]): 44 minutes

2. Calculate the number of half-lives that have passed:
The number of half-lives ([tex]\( n \)[/tex]) is given by:
[tex]\[ n = \frac{t}{t_{1/2}} \][/tex]
Substituting the given values:
[tex]\[ n = \frac{44 \text{ min}}{22 \text{ min}} = 2 \][/tex]
Hence, 2 half-lives have passed.

3. Calculate the remaining mass:
The remaining mass of a substance after a certain number of half-lives can be determined using the formula:
[tex]\[ m = m_0 \times (0.5)^n \][/tex]
Substituting the values:
[tex]\[ m = 7.2 \text{ g} \times (0.5)^2 \][/tex]

4. Perform the calculation:
First, calculate [tex]\( (0.5)^2 \)[/tex]:
[tex]\[ (0.5)^2 = 0.25 \][/tex]
Then multiply this by the initial mass:
[tex]\[ m = 7.2 \text{ g} \times 0.25 = 1.8 \text{ g} \][/tex]

Therefore, the remaining mass of potassium-44 in the body after 44 minutes is 1.8 grams.
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