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A nuclide has a decay rate of \(2.00 \times 10^{10} \mathrm{s}^{-1} .\) After 25.0 days, its decay rate is \(6.25 \times 10^{8} \mathrm{s}^{-1}\). What is the nuclide's half-life? (a) 25.0 d; (b) 12.5 d; (c) 50.0 d; (d) \(5.00 \mathrm{d} ;\) (e) none of these.

Short Answer

Expert verified
The half-life of the nuclide is approximately 6.63 days. The correct answer is (e) none of these.

Step by step solution

01

Identify the Knowns

The initial decay rate \(N_0\) is \(2.00 \times 10^{10} \mathrm{s}^{-1}\). The decay rate after 25 days, denoted \(N\), is \(6.25 \times 10^{8} \mathrm{s}^{-1}\). The time \(t\) is 25 days, but in seconds is \(25 \times 24 \times 60 \times 60 = 2160000 \mathrm{s}\). This converts the time into seconds to match the decay rate units.
02

Use the Exponential Decay Formula

The exponential decay formula is \( N = N_0 \times \exp(-\lambda t)\), where \(N_0\) and \(N\) are initial and final decay rates, \(t\) is time, and \(\lambda\) is the decay constant. We need to solve for \(\lambda\) using the given values: \(6.25 \times 10^{8} = 2.00 \times 10^{10} \times \exp(-\lambda \times 2160000)\).
03

Solve for the Decay Constant

Rearranging the formula and taking the natural logarithm of both sides gives: \(\lambda = - \frac{\ln(6.25 \times 10^{8} / 2.00 \times 10^{10})}{2160000}\). When calculated, this gives \(\lambda = 1.21 \times 10^{-6} \mathrm{s}^{-1}\).
04

Derive Half-Life

The half-life \(T_{1/2}\) is related to the decay constant by \(T_{1/2} = \frac{0.693}{\lambda}\). Substituting the value of \(\lambda\) gives us \(T_{1/2} = \frac{0.693}{1.21 \times 10^{-6}}\). When calculated, this gives \(T_{1/2} = 572600 \mathrm{s}\).
05

Convert Half-Life into Days

Finally, convert the half-life into days: \(T_{1/2} = \frac{572600}{24 \times 60 \times 60}\). When calculated, this gives \(T_{1/2} = 6.63\) days. So the correct answer is not among the options given.

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Key Concepts

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

Exponential Decay
Exponential decay is a fundamental concept in understanding how quantities decrease over time, particularly in processes involving radioactive decay. In essence, it describes how a certain quantity diminishes at a rate proportional to its current value. This behavior is characterized by the exponential decay formula:
  • \( N = N_0 \times \exp(-\lambda t) \),
  • where \( N_0 \) is the initial amount, \( N \) is the amount at time \( t \), \( \lambda \) is the decay constant, and \( \exp \) denotes the exponential function.
This equation implies that the decrease is rapid initially and slows down over time, making it a vital tool in calculating how much of a substance will remain in a given period.
The understanding of exponential decay is crucial when studying half-life in physics or chemistry, as it gives insight into the time-dependent behavior of decaying radioactive elements.
Decay Constant
The decay constant, often symbolized as \( \lambda \), plays a vital role in the characterization of decaying processes, such as the breakdown of radioactive substances. It essentially represents the probability per unit time that a molecule will decay. If you know the decay constant, you can determine how fast or slow a particular substance will decrease over time.
Calculated using the formula:
  • \( \lambda = - \frac{\ln(N/N_0)}{t} \),
  • where \( N \) is the decay rate at time \( t \), \( N_0 \) is the initial decay rate, and \( \ln \) is the natural logarithm.
A larger value of \( \lambda \) indicates a faster decay process, as the substance transitions quickly from its initial to its given state, whereas a smaller decay constant means more gradual decay. Knowing the decay constant aids in deriving the half-life, another critical aspect of radioactive decay analysis.
Nuclide Decay Rate
The nuclide decay rate concerns how fast a particular nuclide, which is a distinct kind of atom or nucleus characterized by a specific number of protons and neutrons, decays over time. Measured in terms of disintegrations per second, this rate is pivotal in nuclear physics and applications like dating geological samples or understanding nuclear reactions.
To comprehend nuclide decay, consider these key points:
  • The initial decay rate \( (N_0) \) provides a benchmark of how rapidly a nuclide starts to decay.
  • Over time, the decay rate \( N \) changes, typically decreasing as nuclides break down into stable forms.
Using the exponential decay formula \( N = N_0 \times \exp(-\lambda t) \), one can calculate how much of a nuclide remains at any given time. Understanding the nuclide decay rate is essential for predicting the behavior of radioactive substances, hence it serves an important function in scientific and engineering calculations.

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Most popular questions from this chapter

One of the following elements has eight naturally occurring stable isotopes. We should expect that one to be (a) \(\mathrm{Ra} ;\) (b) \(\mathrm{Au} ;\) (c) \(\mathrm{Cd} ;\) (d) Br.

Given a radioactive nuclide with \(t_{1 / 2}=1.00 \mathrm{h}\) and a current disintegration rate of 1000 atoms \(s^{-1}\), three hours from now the disintegration rate will be (a) 1000 atoms \(s^{-1} ;\) (b) 333 atoms \(s^{-1} ;\) (c) 250 atoms \(s^{-1}\); (d) 125 atoms \(s^{-1}\)

Briefly describe each of the following ideas, phenomena, or methods: (a) radioactive decay series;(b) charged-particle accelerator; (c) neutron-to- proton ratio; (d) mass-energy relationship; (e) background radiation.

An ester forms from a carboxylic acid and an alcohol. $$\mathrm{RCO}_{2} \mathrm{H}+\mathrm{HOR}^{\prime} \longrightarrow \mathrm{RCO}_{2} \mathrm{R}^{\prime}+\mathrm{H}_{2} \mathrm{O}.$$ This reaction is superficially similar to the reaction of an acid with a base such as sodium hydroxide. The mechanism of the reaction can be followed by using the tracer \(^{18} \mathrm{O}\). This isotope is not radioactive, but other physical measurements can be used to detect its presence. When the esterifcation reaction is carried out with the alcohol containing oxygen-18 atoms, no oxygen-18 beyond its naturally occurring abundance is found in the water produced. How does this result affect the perception that this reaction is like an acid-base reaction?

Write a plausible equation for the decay of tritium, 3 \(\mathrm{H}\), the radioactive isotope of hydrogen. 1 \(\textrm{ }\).

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