/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A source contains two phosphorus... [FREE SOLUTION] | 91Ó°ÊÓ

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A source contains two phosphorus radionuclides, \({ }^{32} \mathrm{P}\left(T_{1 / 2}=\right.\) \(14.3 \mathrm{~d}\) ) and \({ }^{33} \mathrm{P}\left(T_{1 / 2}=25.3 \mathrm{~d}\right.\) ). Initially, \(10.0 \%\) of the decays come from \({ }^{33} \mathrm{P}\). How long must one wait until \(90.0 \%\) do so?

Short Answer

Expert verified
Approximately 49.54 days.

Step by step solution

01

Define Initial Activity Ratio

Initially, 10% of the decays are from \(^{33}\text{P}\), so its ratio of decay to the total decay is \(0.1\) and the ratio of \(^{32}\text{P}\) is \(0.9\). Hence, the activity ratio \( R_{0} = \frac{A_{33,0}}{A_{32,0}} = \frac{0.1}{0.9} = \frac{1}{9} \).
02

Use Decay Formula for Activity Change With Time

The activity of a radionuclide changes over time according to \( A = A_{0} e^{-\lambda t} \), where \( \lambda = \frac{\ln(2)}{T_{1/2}} \). Calculate \( \lambda_{32} = \frac{\ln(2)}{14.3} \) and \( \lambda_{33} = \frac{\ln(2)}{25.3} \). Substitute these values into the decay formulas.
03

Establish Future Activity Ratio

In future, 90% of decays come from \(^{33}\text{P}\), so \( R_{f} = \frac{A_{33}}{A_{32}} = \frac{0.9}{0.1} = 9 \). This is our final desired activity ratio.
04

Set Up Equation for Time of Desired Ratio

Using the decay formulas, express the future activities as \( A_{33} = A_{33,0} e^{-\lambda_{33} t} \) and \( A_{32} = A_{32,0} e^{-\lambda_{32} t} \). The desired ratio leads to \[ \frac{A_{33,0} e^{-\lambda_{33} t}}{A_{32,0} e^{-\lambda_{32} t}} = 9\].
05

Solve for Time t

From the equation \( \frac{1}{9} e^{(\lambda_{32} - \lambda_{33})t} = 9 \), solve for \( t \) by taking natural logarithm on both sides:\( -2 \ln(9) = (\lambda_{32} - \lambda_{33}) t \). Then, calculate \( t = \frac{-2 \ln(9)}{\lambda_{32} - \lambda_{33}} \). Substitute calculated decay constants to find \( t \).
06

Calculate t Numerically

Substitute the numerical values from earlier steps into the equation: \( \lambda_{32} \approx 0.048507 \), \( \lambda_{33} \approx 0.027389 \), to find \( t = \frac{-2 \ln(9)}{0.048507 - 0.027389} \). Compute \( t \approx 49.54 \text{ days} \).

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Key Concepts

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

Half-life
Half-life is the time required for half of the radioactive nuclei in a sample to decay. Every radionuclide has a characteristic half-life which dictates how quickly or slowly it decays. This concept allows us to predict how long it will take for a certain amount of a radionuclide to decay into its daughter products. The half-life is denoted by the symbol \( T_{1/2} \).
Understanding this allows us to derive the decay constant \( \lambda \), which is a measure of the probability per unit time that a nucleus will decay. The decay constant is key in calculating the remaining activity of a sample over time.
  • For a given half-life \( T_{1/2} \), the decay constant is calculated as \( \lambda = \frac{\ln(2)}{T_{1/2}} \).
  • This relationship highlights how radionuclides with shorter half-lives decay more quickly than those with longer half-lives.
Calculating the activity of a radionuclide is thus heavily reliant on the understanding of its half-life and, by extension, the decay constant.
Phosphorus Radionuclides
Phosphorus radionuclides, such as \(^{32}\text{P} \) and \(^{33}\text{P} \), are isotopes of phosphorus that are unstable due to their nuclear configuration.
These isotopes are widely used in biological and medical research, particularly in the study of DNA and RNA as they can be used to label molecules via radioactive tags.
  • \(^{32}\text{P} \) has a half-life of 14.3 days and it is a beta-emitter.
  • \(^{33}\text{P} \) has a longer half-life of 25.3 days and is also a beta-emitter.
Due to their different half-lives, \(^{32}\text{P} \) and \(^{33}\text{P} \) decay at different rates, affecting how long each isotope remains active and useful for labeling purposes in experiments.
The choice between using \(^{32}\text{P} \) or \(^{33}\text{P} \) in experiments largely depends on how long the radioactive labeling needs to remain active.
Activity Ratio
Activity ratio refers to the proportion of decay events attributed to one radionuclide compared to another within a mixture. This ratio changes over time because each radionuclide in a mixture can have a different decay rate.
Initially, in a mixture of \(^{32}\text{P} \) and \(^{33}\text{P} \), you might start with a ratio such as \( 0.1 \) for \(^{33}\text{P} \) and \( 0.9 \) for \(^{32}\text{P} \).
  • This translates to only 10% of initial decays coming from \(^{33}\text{P} \) and the rest from \(^{32}\text{P} \).
  • Over time, as \(^{32}\text{P} \), with its shorter half-life, decays faster, the activity ratio shifts, potentially increasing the proportion of \(^{33}\text{P} \) decays.
The activity ratio is crucial for determining how the decay dynamics affect the overall radiation of a sample over time. It is essential to calculate changes in activity ratios to understand how they influence experimental results, especially in scenarios where the isotopic composition directly affects the outcome of a study.

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Most popular questions from this chapter

The half-life of a radioactive isotope is 140 d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

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What is the binding energy per nucleon of the americium isotope \({ }_{95}^{244} \mathrm{Am} ?\) Here are some atomic masses and the neutron mass. $$\begin{array}{lr}{ }_{95}^{244} \mathrm{Am} & 244.064279 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

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