Question

If the amount of radioactive phosphorus-32 in a sample decreases from \(1.2 \mathrm{mg}\) to \(0.30 \mathrm{mg}\) in 28 days, what is the halflife, in days, of phosphorus- \(32 ?\)

Short answer

The half-life of phosphorus-32 is 14 days.

Step-by-step solution

Understand the problem

We need to find the half-life of phosphorus-32 given that the amount decreases from 1.2 mg to 0.30 mg in 28 days.

Use the decay formula

The general formula for radioactive decay is where: is the final amount, is the initial amount, is the decay constant, and is time.

Substitute known values

Given: days, mg, mg.

Solve for the decay constant

Using the values from Step 3: mg = mg e^{- t} 0.30 mg = 1.2 mg e^{- (28 days)} .

Isolate in the expression

Divide both sides by 1.2 mg: 0.30 / 1.2 = e^{- t}. 0.25 = e^{- (28 )}

Apply the natural logarithm

Taking the of both sides we get: ln(0.25) = ln( e^{- (28)} ) ln(0.25) = - . 28 .

Solve for the decay constant

= - ln(0.25) / 28. Solving we get = 0.0496.

Find the halflife using the decay constant

The halflife ( T_{1/2} ) is given by the expression: T_{1/2} = ln(2) / and = ln(2) / 0.0496 = 14 days.

Key concepts

half-life calculation

To understand half-life calculations, let's start with what a half-life is. The half-life of a radioactive substance is the time it takes for half of the sample to decay. In other words, if you start with 1 mg of a radioactive substance, after one half-life, you will have 0.5 mg left. After another half-life, you will have 0.25 mg left, and so on. Calculating the half-life involves understanding the decay process and using an appropriate mathematical formula.

For phosphorus-32, the problem states that the amount decreases from 1.2 mg to 0.30 mg in 28 days. We can use the decay formula:
\[ N(t) = N_0 e^{-\lambda t} \]
Where:

• \(N(t)\) is the final amount, in this case, 0.30 mg
  
• \(N_0\) is the initial amount, in this case, 1.2 mg
  
• \(\lambda\) is the decay constant
  
• \(t\) is the time, in this case, 28 days
  

By substituting the known values, we can solve for the decay constant \(\lambda\), and then use it to find the half-life. The formula for half-life is:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Using this process, we found the half-life of phosphorus-32 to be 14 days.

decay constant

The decay constant, denoted as \(\lambda\), is a crucial parameter in radioactive decay. It represents the probability that an atom will decay per unit time. A higher decay constant means that the substance decays more quickly.

To find the decay constant, we start with the decay formula:
\[0.30 = 1.2 e^{-\lambda \times 28} \]
We simplify the equation by dividing both sides by 1.2:

\[ \frac{0.30}{1.2} = e^{-\lambda \times 28} \]
which gives:
\[ e^{-\lambda \times 28} = 0.25 \]
Taking the natural logarithm of both sides, we get:
\[ \ln(0.25) = -\lambda \times 28 \]
Solving for \(\lambda\), we have:
\[ \lambda = - \frac{\ln(0.25)}{28} \]
Plugging in the value of \(\ln(0.25)\), which is approximately -1.3863, we find:
\[ \lambda = \frac{1.3863}{28} \approx 0.0496 \]
This is the decay constant for phosphorus-32 in this case.

Knowing \(\lambda\), we can also find the half-life using the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]

phosphorus-32

Phosphorus-32, or \(^{32}P\), is a radioactive isotope commonly used in scientific research and medical treatments. It undergoes beta decay by emitting a beta particle, which is a high-energy electron.

Here are some key points about phosphorus-32:

• Phosphorus-32 has a half-life of about 14 days, which means that it takes 14 days for half of any amount of \(^{32}P\) to decay.
  
• Its primary use is in molecular biology as a tracer because it can label DNA and RNA.
  
• It's also used in cancer treatments, particularly for certain types of leukemia and lymphoma, as it can target rapidly dividing cells.
  

Understanding the behavior of phosphorus-32 helps researchers utilize its properties for beneficial applications.
To recap, the decay constant and half-life both offer key insights into how long phosphorus-32 remains active in any given sample. This knowledge is essential for safely and effectively using it in both scientific and medical contexts.