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What is the half-life of iron- 59 , a radioisotope used medically in the diagnosis of anemia, if a sample with an initial decay rate of 16,800 disintegrations/min decays at a rate of 10,860 disintegrations/min after \(28.0\) days?

Short Answer

Expert verified
The half-life of iron-59 is approximately 44.4 days.

Step by step solution

01

Understand the Problem

We need to find the half-life of iron-59 based on the change in its decay rate over 28 days. The initial decay rate is 16,800 disintegrations/min, and it decreases to 10,860 disintegrations/min.
02

Apply the Decay Formula

The decay formula is \( N(t) = N_0 \times e^{-kt} \), where \( N(t) \) is the remaining amount, \( N_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time.
03

Use Given Values

For the given exercise, \( N_0 = 16,800 \), \( N(t) = 10,860 \), and \( t = 28 \) days. Substitute these in the equation: \( 10,860 = 16,800 \times e^{-28k} \).
04

Solve for Decay Constant \( k \)

Divide both sides by 16,800: \( \frac{10,860}{16,800} = e^{-28k} \). Calculate: \( 0.6464 = e^{-28k} \). Take the natural logarithm of both sides: \( \ln(0.6464) = -28k \).
05

Calculate \( k \)

Using a calculator: \( \ln(0.6464) \approx -0.4365 \), so \( k = \frac{-0.4365}{-28} \approx 0.01559 \text{ day}^{-1} \).
06

Calculate Half-life

The half-life is given by \( t_{1/2} = \frac{\ln(2)}{k} \). Compute: \( t_{1/2} = \frac{0.693}{0.01559} \).
07

Compute the Final Answer

After calculating, \( t_{1/2} \approx 44.44 \) days. Thus, the half-life of iron-59 is approximately 44.4 days.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
Half-life is a fascinating concept in radioactive decay, which refers to the time taken for half of a radioactive substance to decay into another substance. It is a crucial measurement in understanding the behavior of radioactive materials.

The half-life of a radioactive element can be calculated using the formula:
  • \( t_{1/2} = \frac{\ln(2)}{k} \)
where \( t_{1/2} \) is the half-life, and \( k \) is the decay constant. Naturally, this formula stems from the exponential decay formula

It's important to note that the half-life is constant and independent of the amount of substance you have. This means that whether you have a gram, a kilogram, or a ton of the substance, the time it takes for half to decay remains the same. In practical terms, this allows scientists and medical professionals to predict how quickly a sample of radioactive material will decrease in activity over time.
Iron-59
Iron-59 is a radioisotope commonly used in the medical field, particularly in the diagnosis of anemia. It works by tagging red blood cells, allowing physicians to track the cells' production and destruction within the body. This is crucial for diagnosing conditions where there is abnormal production or destruction of blood cells.

The radioactivity of iron-59 is helpful because:
  • It has a suitably long half-life, making it viable for detailed medical studies.
  • It emits detectable gamma rays, which can be measured externally without requiring invasive procedures.
Despite its benefits, handling iron-59 requires appropriate safety precautions due to its radioactive nature. Proper precautions ensure safety for both the medical staff and the patients.
Decay Constant
The decay constant, represented by \( k \), is a measure of the probability that a given atom will decay per unit time. It's an essential parameter in the study of radioactive substances because it helps quantify the rate at which a substance undergoes decay.

To find the decay constant, we use the formula:
  • \( N(t) = N_0 \times e^{-kt} \)
where \( N(t) \) is the amount of substance remaining after time \( t \), \( N_0 \) is the initial amount, and \( t \) is the time elapsed.

The decay constant can be found once you know the initial and final quantities of the substance and the time over which decay has occurred. In the case of iron-59, you would observe the change in the number of disintegrations per minute over a specific time period, such as 28 days, to find \( k \). Once determined, \( k \) can be used to understand the decay characteristics and further calculate the half-life of the radioactive isotope.

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Most popular questions from this chapter

The age of any remains from a once-living organism can be determined by radiocarbon dating, a procedure that works by determining the concentration of radioactive \({ }^{14} \mathrm{C}\) in the remains. All living organisms contain an equilibrium concentration of radioactive \({ }^{14} \mathrm{C}\) that gives rise to an average of \(15.3\) nuclear decay events per minute per gram of carbon. At death, however, no additional \({ }^{14} \mathrm{C}\) is taken in, so the concentration slowly drops as radioactive decay occurs. What is the age of a bone fragment from an archaeological dig if the bone shows an average of \(2.3\) radioactive events per minute per gram of carbon? For \({ }^{14} \mathrm{C}, t_{1 / 2}=5715\) years.

The age of an igneous rock that has solidified from magma can be found by analyzing the amount of \({ }^{40} \mathrm{~K}\) and 40 Ar. If the rock contains \(1.20 \mathrm{mmol}\) of \({ }^{40} \mathrm{~K}\) and \(0.95 \mathrm{mmol}\) of \({ }^{40} \mathrm{Ar}\), how long ago did the rock cool? The half-life of potassium- 40 is \(1.25 \times 10^{9}\) years. $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar}+{ }_{1}^{0} \mathrm{e} $$

Cesium- 137 is a radioactive isotope released as a result of the Fukushima Daiichi nuclear disaster in Japan in 2011 . If \(89.2 \%\) remains after \(5.00\) years, what is the half-life?

The present level of \({ }^{235} \mathrm{U}\) in naturally occurring uranium ore is \(0.72 \%\). If the half-life of uranium- 235 is \(7.03 \times 10^{8}\) years, how many years ago did naturally occurring uranium contain \(3.00 \%{ }^{235} \mathrm{U}\), the level needed to sustain a chain reaction?

Of the two isotopes of iodine, \({ }^{136} \mathrm{I}\) and \({ }^{122} \mathrm{I}\), one decays by \(\beta\) emission and one decays by positron emission. Which does which? Explain.

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