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Chapter 19: Problem 50

The half-life of indium- 111, a radioisotope used in studying the distribution of white blood cells, is \(t_{1 / 2}=2.805\) days. What is the decay constant of \({ }^{111}\) In?

Short Answer

Expert verified
The decay constant of \( ^{111} \)In is approximately \( 0.247 \text{ day}^{-1} \).

Step by step solution

01

Identify the Known Value

The half-life, denoted as \( t_{1/2} \), of indium-111 is given as \( 2.805 \) days.
02

Use the Half-Life Formula for Decay Constant

The decay constant, \( \lambda \), is related to the half-life by the equation:\[\lambda = \frac{\ln(2)}{t_{1/2}}\]This equation arises because the half-life is the time required for half of the radioactive nuclei to decay.
03

Plug in the Known Half-Life

Substitute the given half-life \( t_{1/2} = 2.805 \) days into the formula:\[\lambda = \frac{\ln(2)}{2.805}\]
04

Calculate the Natural Logarithm

Calculate \( \ln(2) \), which is approximately \( 0.693 \).
05

Perform the Division

Divide \( 0.693 \) by \( 2.805 \) to find the decay constant:\[\lambda \approx \frac{0.693}{2.805} \approx 0.247 \text{ day}^{-1}\]

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Key Concepts

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

half-life
Half-life is a fundamental concept in the study of radioactive decay. It refers to the time it takes for half of the radioactive nuclei in a sample to decay into its products. In the context of indium-111, the half-life is given as 2.805 days. This value tells us how long it takes for half of the indium-111 atoms to transform.

Understanding half-life is crucial because it helps scientists predict how long a radioactive element will remain active. Knowing the half-life can also enable us to calculate the decay constant, a key parameter that describes the rate of decay.
  • Half-life is specific to each radioactive isotope.
  • It provides insights into how quickly a substance will lose its radioactivity.
  • The half-life remains constant regardless of the starting quantity of the sample.
decay constant
The decay constant, denoted by the symbol \( \lambda \), is a measure that describes the speed of radioactive decay. It is related to the half-life of a substance. The smaller the decay constant, the slower the decay rate. For indium-111, we calculated the decay constant using its half-life.

To find the decay constant using half-life, we use the formula:\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
This equation shows the relationship between decay constant and half-life, where \( \ln(2) \approx 0.693 \) is the natural logarithm of 2. The decay constant tells how quickly the indium-111 will reduce by half in its original quantity.
  • The decay constant is expressed in units of inverse time, like \( \text{day}^{-1} \).
  • It is a crucial factor in the exponential decay equation.
  • Understanding \( \lambda \) helps in calculating the remaining quantity of a substance over time.
natural logarithm
The natural logarithm, represented by \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is used extensively in exponential growth and decay problems, including radioactive decay. When computing the decay constant from half-life, the natural log of 2, \( \ln(2) \), plays a crucial role.

In our calculations for indium-111, we used \( \ln(2) \) because it is the factor that transforms the equation from describing half-life to expressing the decay constant. Mathematically, the value of \( \ln(2) \) is approximately 0.693, and it is used in the decay constant calculation formula:\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
  • \( \ln \) helps convert complex exponential relationships into simpler linear ones.
  • It allows easier manipulation in calculus and algebra.
  • Understanding \( \ln(2) \) is essential for working with half-life and decay constant equations.

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Most popular questions from this chapter

Radon- 222 decays by a series of three \(\alpha\) emissions and two \(\beta\) emissions. What is the final stable nucleus?

Uranium- 238 undergoes alpha decay to thorium- \(234 .\) (a) Write a balanced nuclear equation. (b) Calculate the mass change in \((\mathrm{g} /\) atom \()\) and energy change in \((\mathrm{kJ} / \mathrm{mol})\). (c) Is energy absorbed or released when uranium undergoes radioactive decay? The masses are \({ }^{238} \mathrm{U}(238.0508 \mathrm{u}),{ }^{234} \mathrm{Th}(234.0436 \mathrm{u}),{ }^{4} \mathrm{He}(4.0026 \mathrm{u})\).

The radioactive isotope \({ }^{100} \mathrm{~T}_{c}\) decays to form the stable isotope \({ }^{100}\) Mo. (a) There are two possible pathways for this decay. Write balanced equations for both. (b) Only one of the pathways is observed. Calculate the energy released by both pathways, and explain why only one is observed. Relevant masses are: \({ }^{100} \mathrm{~T} c(99.907657 \mathrm{u}),{ }^{100} \mathrm{Mo}\) \((99.90748 \mathrm{u}), \mathrm{e}^{-}(0.0005486 \mathrm{u})\)

In an endothermic chemical reaction, do the products have more mass, less mass, or the same mass as the reactants? Explain.

Polonium- 210 , a naturally occurring radioisotope, is an \(\alpha\) emitter, with \(t_{1 / 2}=138 \mathrm{~d}\). Assume that a sample of \({ }^{210} \mathrm{Po}\) with a mass of \(0.700 \mathrm{mg}\) was placed in a \(250.0-\mathrm{mL}\) flask, which was evacuated, sealed, and allowed to sit undisturbed. What would the pressure be inside the flask (in \(\mathrm{mm} \mathrm{Hg}\) ) at \(20^{\circ} \mathrm{C}\) after 365 days if all the \(\alpha\) particles emitted had become helium atoms?
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