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A freshly isolated sample of \({ }^{90} \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1:00 P.M. on December 3, 2006. At 2:15 P.M. on December \(17,2006,\) its activity was measured again and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \({ }^{90} \mathrm{Y}.\)

Short Answer

Expert verified
The half-life of \(^{90} Y\) is approximately 64.35 hours.

Step by step solution

01

Determine the Time Interval

Find the total time between the two measurements from 1:00 PM on December 3, 2006, to 2:15 PM on December 17, 2006. Break down the calculation: 1. From 1:00 PM December 3 to 1:00 PM December 17 is 14 days. 2. From 1:00 PM to 2:15 PM on December 17 adds another 1.25 hours. Thus, total time = 14 days * 24 hours/day + 1.25 hours = 337.25 hours.
02

Use the Decay Formula

The radioactive decay formula is: \[ A = A_0 \times e^{-\lambda t} \]where:- \( A \) is the final activity, \( 2.6 \times 10^4 \) disintegrations per minute.- \( A_0 \) is the original activity, \( 9.8 \times 10^5 \) disintegrations per minute.- \( \lambda \) is the decay constant.- \( t \) is the time in hours, 337.25.Rearrange to find \( \lambda \):\[ \lambda = \frac{1}{t} \times \ln\left(\frac{A_0}{A}\right) \]
03

Calculate the Decay Constant

Substitute the known values into the formula:\[ \lambda = \frac{1}{337.25} \times \ln\left(\frac{9.8 \times 10^5}{2.6 \times 10^4}\right) \]Calculate:\[ \frac{9.8 \times 10^5}{2.6 \times 10^4} \approx 37.6923 \]\[ \ln(37.6923) \approx 3.6326 \]\[ \lambda \approx \frac{3.6326}{337.25} \approx 0.010768 \] (per hour)
04

Calculate the Half-Life

Half-life \( T_{1/2} \) is related to the decay constant by the formula:\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]Substitute the value of \( \lambda \):\[ T_{1/2} = \frac{0.693}{0.010768} \approx 64.35 \] hours.

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Key Concepts

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

Half-life calculation
Radioactive substances decay exponentially over time. A crucial concept in understanding this process is the half-life. The half-life (\( T_{1/2} \) ) of a radioactive element is the time it takes for half of a sample to decay. This property is unique to each radioactive isotope and remains constant.
To find the half-life, we use the decay constant (\( \lambda \)), which describes the rate of particle decay. The relationship between these two quantities is expressed in the formula: \[ T_{1/2} = \frac{\ln(2)}{\lambda} \]This formula tells us how long it will take for half of the material to decay, providing a predictable measure of stability. In practice, knowing the half-life helps in various applications, from dating archaeological finds to medical diagnostics. Keep in mind that the exponent \( \ln(2) \) corresponds to the natural logarithm of 2, approximating 0.693.
Decay constant
The decay constant, denoted as \( \lambda \), is a fundamental component in the study of radioactivity. It represents the probability of a single atom decaying per unit time, serving as an indicator of how quickly or slowly a radioactive substance will degrade.
To calculate the decay constant, we rearrange the radioactive decay equation:\[ \lambda = \frac{1}{t} \times \ln\left(\frac{A_0}{A}\right) \]where \( A_0 \) is the initial activity, \( A \) is the activity at time \( t \), and \( t \) is the elapsed time.
  • The higher the decay constant, the faster the decay rate.
  • Conversely, a smaller decay constant indicates a slower decay rate, leading to a longer half-life.
Understanding \( \lambda \) allows scientists to predict how the activity of a radioactive sample changes over time, crucial in research and safety measures.
Radioactivity
Radioactivity refers to the spontaneous disintegration of atomic nuclei, leading to the emission of particles or electromagnetic radiation. This natural process is the basis of radioactive decay.
Elements exhibit radioactivity when they have unstable isotopes. Over time, these isotopes seek stability by transforming into different elements or isotopes, releasing energy in the form of radiation.
Radioactivity is used in diverse fields:
  • In medicine, it helps diagnose and treat certain conditions.
  • In power generation, it aids in producing nuclear energy.
  • In archaeology, it plays a role in dating ancient artifacts.
Understanding the fundamental nature of radioactivity helps in both harnessing its benefits and safeguarding against its potential dangers.
Disintegration rate
The disintegration rate, or activity, of a radioactive material reflects how many atoms decay over a specific timeframe. It is measured in disintegrations per unit time, often per second or per minute.
This rate helps quantify the intensity of a radioactive sample and is influenced by both the number of unstable atoms present and the decay constant. For instance, if a sample of \({ }^{90} \mathrm{Y} \) originally has an activity of \(9.8 \times 10^5\) disintegrations per minute which later becomes \(2.6 \times 10^4\), the decrease in activity over time reflects the ongoing decay process.
  • High disintegration rates imply a large number of decaying atoms, hence a more active sample.
  • Low disintegration rates indicate fewer atoms decaying, resulting in a more stable lifecycle.
The disintegration rate is crucial for accurate assessments in fields like nuclear medicine, environmental safety, and research.

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Most popular questions from this chapter

Cobalt- 60 is an isotope used in diagnostic medicine and cancer treatment. It decays with \(\gamma\) -ray emission. Calculate the wavelength of the radiation in nanometers if the energy of the \(\gamma\) ray is \(2.4 \times 10^{-13} \mathrm{~J} / \mathrm{photon} .\)

Explain why achievement of nuclear fusion in the laboratory requires a temperature of about 100 million degrees Celsius, which is much higher than that in the interior of the sun ( 15 million degrees Celsius).

Describe how you would use a radioactive iodine isotope to demonstrate that the following process is in dynamic equilibrium: $$ \mathrm{PbI}_{2}(s) \rightleftarrows \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$

A radioactive isotope of copper decays as follows: $$ { }^{64} \mathrm{Cu} \longrightarrow{ }^{64} \mathrm{Zn}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=12.8 $$ Starting with \(84.0 \mathrm{~g}\) of \({ }^{64} \mathrm{Cu},\) calculate the quantity of \({ }^{64} \mathrm{Zn}\) produced after \(18.4 \mathrm{~h} .\)

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}(89.907738 \mathrm{amu})\) : $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: \({ }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{1 / 2}=64 \mathrm{~h}\) Zirconium- \(90(89.904703 \mathrm{amu})\) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times 10^{-4}\) amu.) (b) Starting with 1 mole of \({ }^{90} \mathrm{Sr}\), calculate the number of moles of \({ }^{90} \mathrm{Sr}\) that will decay in a year. (c) Calculate the amount of heat released (in kJ) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in part \((\mathrm{b}).\)

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