The decay constant of \({ }_{6} \mathrm{C}^{14}\) is \(2.31 \times 10^{-4}\) year \(^{-1}\). Its half life is: (a) \(2 \times 10^{3}\) yrs (b) \(2.5 \times 10^{3}\) yrs (c) \(3 \times 10^{3}\) yrs (d) \(3.5 \times 10^{3} \mathrm{yrs}\)

Short Answer

Expert verified
The half-life of \( _{6} \mathrm{C}^{14} \) is approximately \(3 \times 10^{3}\) years, which corresponds to option (c).

Step by step solution

01

Understanding the Relationship between Decay Constant and Half-life

The decay constant (\(\lambda\)) and half-life (\(t_{1/2}\)) are inversely proportional to each other and their relationship is defined by the equation \(t_{1/2} = \frac{\ln{2}}{\lambda}\). Knowing the decay constant, we can calculate the half-life using this formula.
02

Calculating the Half-life

Substitute the decay constant value \(\lambda = 2.31 \times 10^{-4} \) year\(^{-1}\) into the half-life formula: \(t_{1/2} = \frac{\ln{2}}{2.31 \times 10^{-4}}\). Use a calculator to find the value of \(\ln{2}\), which is approximately 0.693.
03

Performing the Calculation

Perform the calculation to find the half-life: \(t_{1/2} = \frac{0.693}{2.31 \times 10^{-4}}\) which equals approximately \(3000\) years.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Decay Constant
The concept of the decay constant, symbolized by \(\lambda\), is crucial in understanding radioactive decay. It represents the probability per unit time of a single atom decaying. One might wonder how it's possible to assign a time value to the decay of individual atoms, which seem to act unpredictably. Indeed, it is unpredictable for one atom, but when we scale up to a sample containing a large number of atoms, the average behavior becomes remarkably consistent and can be described mathematically.

The decay constant is unique to each radioactive isotope and essentially sets the pace at which the isotope will decay. A larger decay constant indicates that the isotope will decay more quickly, whereas a smaller value suggests that the atoms will remain unchanged for longer periods. In mathematical terms, if one has the decay constant value, one can employ it to calculate not only the half-life of the isotope but also to predict the remaining quantity of a substance after a certain period using the exponential decay formula.
Radioactive Isotope Decay
Radioactive isotope decay is the process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This transformation often results in the nucleus changing to a different element or a different isotope of the same element.

Understanding the decay process is important for a variety of applications, including radioactive dating techniques, medical treatments like cancer therapy, and ensuring safety around nuclear materials. What essentially happens during decay is that the unstable isotope seeks a more stable configuration.

Examples in Nature and Technology

In nature, carbon-14 (\( _{6}\mathrm{C}^{14} \)) decays to nitrogen-14, while in technology, uranium-235 is used in nuclear reactors and atomic bombs due to its propensity to undergo fission when its nucleus decays. These decays happen at rates defined by each isotope's decay constant.
Exponential Decay in Chemistry
Exponential decay is a fundamental concept in chemistry that describes the decrease in the amount of a substance as time progresses. It is exponential because the quantity of the substance does not decrease by a constant amount each period; rather, it decreases by a constant proportion of its current value. This pattern of decay is different from linear decay, where the amount would decrease by the same absolute amount over each time interval.

The formula describing exponential decay is \( N(t) = N_0 e^{-\lambda t} \) where \( N(t) \) represents the remaining quantity of the substance at time \( t \), \( N_0 \) is the initial quantity, \( \lambda \) is the decay constant, and \( e \) is the base of the natural logarithm. When applied, this formula can predict how much of a radioactive substance remains after a given period, which is vital in fields like pharmacokinetics, where it is used to model drug elimination from the body.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following statements is correct? (1) Order of a reaction can be known from experimental results and not from the stoichiometry of a reaction. (2) Molecularity a reaction refers to (i) each of the elementary steps in (an overall mechanism of) a complex reaction or (ii) a single step reaction. (3) Overall molecularity of a reaction may be determined in a manner similar to overall order of reaction. (4) Overall order of a reaction \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{\mathrm{n}} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) is \(\mathrm{m}+\mathrm{n}\) Select the correct answer using the following codes: (a) 2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 3

A catalyst is a substance which: (a) Supplies energy to the reaction (b) Increases the equilibrium concentration of the product (c) Changes the equilibrium constant of the reaction (d) Shortens the time to each equilibrium

The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, hi 40 minutes, the concentration of \(X\) changes from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\), then the rate of reaction when concentration of \(\mathrm{X}\) is \(0.01 \mathrm{M}\) is: (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)

The experimental rate law for a reaction: \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow\) Product, is \(\mathrm{V} \propto \mathrm{C}_{\Lambda} \mathrm{C}_{\mathrm{B}}^{1 / 2} .\) If the concentration of both \(\mathrm{A}\) and are doubled the rate of reaction increases by a factor of (a) \(\sqrt{2}\) (b) 2 (c) \(2 . \sqrt{2}\) (d) 4

A graph plotted between concentration of reactant, consumed at any time ( \(\mathrm{x}\) ) and time ' \(\mathrm{t}\) ' is found to be a straight line passing through the origin. The reaction is of: (a) First-order (b) Zero-order (c) Third-order (d) Second-order

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free