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The rate constant for a certain radioactive nuclide is \(1.0 \times 10^{-3} \mathrm{~h}^{-1}\). What is the half-life of this nuclide?

Short Answer

Expert verified
The half-life (T鈧/鈧) of the nuclide can be calculated using the relation \(T_{1/2} = \frac{ln(2)}{k}\), where k is the given rate constant \(k = 1.0 \times 10^{-3} \mathrm{h}^{-1}\). Plugging in the given value for k, we find that the half-life of the nuclide is approximately \(T_{1/2} \approx 693.1 \mathrm{h}\).

Step by step solution

01

Note down the given values

We are given the rate constant (k) for a radioactive nuclide: \(k = 1.0 \times 10^{-3} \mathrm{h}^{-1}\) Our task is to find the half-life (T鈧/鈧) of the nuclide.
02

Use the formula to find the half-life

Now, we will use the formula that relates the rate constant (k) with half-life (T鈧/鈧): \(T_{1/2} = \frac{ln(2)}{k}\) Here, ln(2) is the natural logarithm of 2. Plug in the given value for k and find the half-life.
03

Calculate the half-life

Insert the given value for k into the formula: \(T_{1/2} = \frac{ln(2)}{1.0 \times 10^{-3} \mathrm{h}^{-1}}\) Calculate the half-life: \(T_{1/2} \approx \frac{0.6931}{1.0 \times 10^{-3} \mathrm{h}^{-1}} = 693.1 \mathrm{h}\) So, the half-life of the nuclide is approximately 693.1 hours.

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Key Concepts

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

Rate Constant
The rate constant is a crucial component when studying radioactive decay or chemical reactions. It essentially represents the speed at which a radioactive material decays or a reaction occurs. For any radioactive nuclide, this constant, denoted as \( k \), is expressed in inverse time units (e.g., \( \mathrm{h}^{-1} \) or \( \mathrm{s}^{-1} \)). This indicates how frequently the decay process proceeds over a unit of time. Understanding the rate constant aids in determining how fast a particular radioactive element will diminish over time. The larger the rate constant, the faster the decay occurs. In our example, the rate constant is \( 1.0 \times 10^{-3} \mathrm{~h}^{-1} \), which helps us calculate essential details like the half-life of the nuclide. Now that you know what a rate constant is, it's easier to see how it directly influences the lifespan of the decaying substance through connected calculations.
Half-Life Calculation
Calculating the half-life of a radioactive nuclide allows us to understand how long it takes for half of a substance to decay. This calculation is vital for applications in fields such as chemistry, physics, medicine, and archeology. The half-life is determined using the following formula:\[ T_{1/2} = \frac{\ln(2)}{k} \]This formula shows that the half-life \( T_{1/2} \) is inversely proportional to the rate constant \( k \). The natural logarithm of 2, \( \ln(2) \approx 0.6931 \), is a constant value used in this equation to reckon the behavior of exponential decay. In the exercise, we find the half-life by plugging the given rate constant \( k = 1.0 \times 10^{-3} \mathrm{h}^{-1} \) into the formula. This computation demonstrates that the nuclide's half-life approximates to 693.1 hours. Half-life is particularly useful in deducing other properties of the decay process, such as predicting how much of a substance remains after a certain period.
Natural Logarithm
Natural logarithms, often denoted as \( \ln \), are logarithms based on the constant \( e \) (approximately 2.71828). They play an important role in many scientific calculations, especially those involving exponential growth or decay, like the behavior of radioactive substances. In the context of radioactive decay, the natural logarithm of 2, represented as \( \ln(2) \approx 0.6931 \), is significant because it relates directly to the concept of half-life. The half-life formula \( T_{1/2} = \frac{\ln(2)}{k} \) makes use of \( \ln(2) \) to accurately convey how exponential decay rhythms correspond to the time for half a substance to vanish.Grasping natural logarithms is essential as they simplify how we calculate and understand the dynamics of various natural processes. Using \( \ln \) in such contexts allows scientists and students alike to gain precise insights into decay or growth phenomena.

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Most popular questions from this chapter

The most significant source of natural radiation is radon- \(222 .\) \({ }^{222} \mathrm{Rn}\), a decay product of \({ }^{238} \mathrm{U}\), is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \({ }^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of \(3.82\) days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \({ }^{238} \mathrm{U}\) decays to \({ }^{222} \mathrm{Rn}\) ? What nuclei are produced when \({ }^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \({ }^{222} \mathrm{Rn}\) ? c. Another problem associated with \({ }^{222} \mathrm{Rn}\) is that the decay of \({ }^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=3.11\right.\) min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) -particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \({ }^{222} \mathrm{Rn}\) levels not exceed \(4 \mathrm{pCi}\) per liter of air \((1 \mathrm{Ci}=\) 1 curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times\) \(10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \({ }^{222} \mathrm{Rn}\) per liter of air.

Technetium-99 has been used as a radiographic agent in bone scans \(\left({ }_{4,}^{99} \mathrm{Tc}\right.\) is absorbed by bones). If \({ }_{43}^{99} \mathrm{Tc}\) has a half-life of \(6.0\) hours, what fraction of an administered dose of \(100 . \mu \mathrm{g}_{43}^{99} \mathrm{Tc}\) remains in a patient's body after \(2.0\) days?

Write balanced equations for each of the processes described below. a. Chromium- 51 , which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- 32, which accumulates in the liver, decays by \(\beta\) particle production.

A small atomic bomb releases energy equivalent to the detonation of 20,000 tons of TNT; a ton of TNT releases \(4 \times 10^{9} \mathrm{~J}\) of energy when exploded. Using \(2 \times 10^{13} \mathrm{~J} / \mathrm{mol}\) as the energy released by fission of \({ }^{235} \mathrm{U}\), approximately what mass of \({ }^{235} \mathrm{U}\) undergoes fission in this atomic bomb?

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \(^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

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