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Determine the half-life of Bk given that a plot of \(\ln N\) against \(t\) is linear with a gradient of -0.0023 day \(^{-1}\) where \(N\) is the number of nuclides present at time \(t\)

Short Answer

Expert verified
The half-life of Bk is approximately 301.30 days.

Step by step solution

01

Understanding the relationship

The relationship between the quantity of a radioactive substance and time can be described by the exponential decay equation: \[N(t) = N_0 e^{-kt}\]where \(N(t)\) is the number at time \(t\), \(N_0\) is the initial amount, and \(k\) is the decay constant. The natural logarithm of this equation is:\[\ln N(t) = \ln N_0 - kt\] This indicates that when plotting \(\ln N\) against \(t\), the slope of the line is \(-k\). From the question, the gradient \(-k = -0.0023 \text{ day}^{-1}\).
02

Relating the gradient to the half-life

The concept of half-life \(t_{1/2}\) is related to the decay constant \(k\) through the equation:\[t_{1/2} = \frac{\ln 2}{k}\]Since we know \(k = 0.0023 \text{ day}^{-1}\), we can substitute it into this equation to find the half-life.
03

Calculating the half-life

Substitute the given decay constant into the half-life equation:\[t_{1/2} = \frac{\ln 2}{0.0023}\approx \frac{0.693}{0.0023} \approx 301.30 \text{ days}\]Thus, the half-life of Bk is approximately 301.30 days.

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Key Concepts

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

Half-Life Calculation
The concept of half-life is crucial when studying radioactive decay. It is the time required for half of the radioactive substance to decay. This period remains constant regardless of how much of the substance is present. An important equation associated with half-life is:\[t_{1/2} = \frac{\ln 2}{k}\]where \( t_{1/2} \) represents the half-life, and \( k \) is the decay constant. The symbol \( \ln 2 \) (approximately 0.693) arises because radioactivity follows an exponential decay process.Understanding how to calculate half-life gives us a way to predict how long a substance will take to reduce to half its initial amount. This calculation is fundamental in fields like nuclear physics, medicine, and environmental science.
Decay Constant
The decay constant \( k \) is a parameter that expresses the rate at which a radioactive substance undergoes decay. It is pivotal in determining both the speed of decay and the half-life of the material. The relationship of decay constant to the decay process can be visualized through the formula:\[N(t) = N_0 e^{-kt}\]Here, \( N(t) \) is the amount of substance left at time \( t \), \( N_0 \) is the initial quantity, and \( e \) is the base of the natural logarithm.Key facts about decay constant:
  • A larger \( k \) indicates a faster decay, meaning the substance will lose half of its mass more quickly.
  • Decay constant is specific to each radioactive material and cannot change under normal conditions.
  • Understanding \( k \) allows scientists to perform predictions and calculations about the lifetime of a radioactive element.
Exponential Decay Equation
The exponential decay equation is fundamental in describing how radioactive materials decay over time. It takes the form:\[N(t) = N_0 e^{-kt}\]This equation signifies that the quantity of radioactive substance decreases exponentially as time progresses. The process reflects that, with each passing unit of time, the substance reduces by a certain percentage, determined by the decay constant \( k \).Some practical applications of the exponential decay equation include:
  • Predicting the behavior of radioactive isotopes in nuclear reactors.
  • Estimating the age of fossils and geological samples through radiocarbon dating.
  • Calculating the dosage of radioactive isotopes in medical treatments.
Natural Logarithm in Decay
In the study of radioactive decay, the natural logarithm (often denoted as \( \ln \)) is central to understanding and simplifying equations. By applying a natural logarithm to the exponential decay equation, we obtain:\[\ln N(t) = \ln N_0 - kt\]This transformation makes it easier to analyze the data by providing a linear relationship when plotting. The slope of the resulting line (\(-k\)) directly relates to the decay constant, aiding in identifying characteristics of the substance's decay rate.In practice, natural logarithms:
  • Allow for straightforward comparison of decay rates between different substances.
  • Simplify computations and problem-solving involving decay processes.
  • Provide clear interpretation of data from decay experiments, making patterns in decay behavior evident.

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Most popular questions from this chapter

Interpret the following notational forms of nuclear reactions: (a) \(\frac{58}{26} \mathrm{Fe}(2 \mathrm{n}, \beta)_{27}^{60} \mathrm{Co} ;(\mathrm{b}) \frac{55}{25} \mathrm{Mn}(\mathrm{n}, \gamma)_{25}^{56} \mathrm{Mn}\) \((\mathrm{c})_{16}^{32} \mathrm{S}(\mathrm{n}, \mathrm{p})_{15}^{32} \mathrm{P}\) (d) 233 11 \(\mathrm{Na}(\gamma, 3 \mathrm{n})_{11}^{20} \mathrm{Na}\)

Account for the fact that the \(^{29}\) Si NMR spectrum of a mixture of \(\mathrm{SiCl}_{4}\) and \(\mathrm{SiBr}_{4}\) that has been standing for \(40 \mathrm{h}\) contains five singlets which include those assigned to \(\mathrm{SiCl}_{4}\) \((\delta-19 \mathrm{ppm})\) and \(\mathrm{SiBr}_{4}(\delta-90 \mathrm{ppm})\)

The \(^{19} \mathrm{F}\) NM \(\mathrm{R}\) spectrum of each of the following molecules exhibits one signal. For which species is this observation consistent with a static molecular structure as predicted by the VSEPR model: (a) \(\mathrm{SiF}_{4}\) i; (b) \(P F_{5} ;\) (c) \(\mathrm{SF}_{6}\) (d) \(\operatorname{SOF}_{2}\) (e) \(\mathrm{CF}_{4} ?\)

Identify the second fission product in the following reactions: (a) \(\frac{235}{92} \mathrm{U}+_{0}^{1} \mathrm{n}--^{142} \mathrm{s} \mathrm{Ba}+?+2_{0}^{1} \mathrm{n}\) (b) \(\frac{235}{92} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow_{52}^{137} \mathrm{Te}+?+2_{0}^{1} \mathrm{n}\)

If the oxide \(P_{4} O_{6}\) is dissolved in an aqueous solution of sodium carbonate, compound A of formula \(\mathrm{Na}_{2} \mathrm{HPO}_{3}\) may be crystallized from solution. The IR spectrum of \(\mathbf{A}\) contains a band at \(2300 \mathrm{cm}^{-1}\). The corresponding band in the IR spectrum of \(\mathbf{B}\) (obtained by an analogous method from \(\mathrm{P}_{4} \mathrm{O}_{6}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) dissolved in \(\mathrm{D}_{2} \mathrm{O}\) ) is at \(1630 \mathrm{cm}^{-1}\) On recrystallization of A from \(\mathrm{D}_{2} \mathrm{O}\), however, its IR spectrum is not affected. Discuss the interpretation of these observations.

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