/*! 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 16 The half-life of an isotope with... [FREE SOLUTION] | 91Ó°ÊÓ

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The half-life of an isotope with a very long half-life cannot be measured by observing its activity as a function of time, since the variation in activity over any reasonable time interval would be too small to be observed. Let \(m\) be the mass in grams of a given isotope of long half-life. (a) Show that the number of nuclei present in this quantity is \(N_{0}=\frac{m}{\mu} N_{A}\) where \(\mu\) is the molar mass of the isotope in \(\mathrm{g} \mathrm{mol}^{-1}\) and \(N_{A}\) is the Avogadro constant. (b) From \(A=-\frac{d N}{d t}=\left(N_{0} \lambda\right) \mathrm{e}^{-\lambda t}\) show that the initial activity is \(A_{0}=\frac{m N_{A}}{\mu} \lambda\) and hence that the half-life can be determined by measuring the initial activity (in Bq) and the mass of the sample (in grams).

Short Answer

Expert verified
The number of nuclei in a sample is given by \(N_0 = \frac{m}{\u\u20mu} N_A\). The initial activity is \(A_0 = \frac{m N_A}{\mu} \lambda\), where \(\lambda\) is the decay constant.

Step by step solution

01

Understand the Problem

An isotope has a very long half-life, making it difficult to measure its activity directly over a short period. We need to show how the number of nuclei in a given mass of this isotope can be calculated and how the initial activity relates to the isotope's half-life.
02

Find the Number of Nuclei

The number of nuclei, denoted as \(N_0\), can be found by using the relationship between mass, molar mass, and the Avogadro constant. The formula given is: \[N_0 = \frac{m}{\mu} N_A\].
03

Explain Number of Nuclei Formula

The formula \(N_0 = \frac{m}{\mu} N_A\) is derived from the fact that \(\mu\) is the molar mass and \(N_A\) is the number of atoms per mole. So for a given mass \(m\) in grams, dividing by the molar mass \(\mu\) gives the number of moles, and multiplying by \(N_A\) converts moles to the number of atoms.
04

Understand Initial Activity

The activity, \(A\), of a radioactive substance is given by \(A = -\frac{dN}{dt} = N_0 \lambda e^{-\lambda t}\), where \(\lambda\) is the decay constant. The initial activity \(A_0\) is the activity at time \(t = 0\).
05

Derive Initial Activity

At \(t = 0\), the formula for activity becomes: \[A_0 = N_0 \lambda = \left(\frac{m}{\mu} N_A \right) \lambda\].
06

Connect Initial Activity and Half-Life

The decay constant \(\lambda\) is related to the half-life \(T_{1/2}\) by the formula \(\lambda = \frac{\ln(2)}{T_{1/2}}\). Therefore, the initial activity depends on the mass of the sample, the Avogadro constant, the molar mass, and the decay constant.
07

Conclude

In conclusion, \(A_0 = \frac{m}{\mu} N_A \lambda \) shows that by measuring the initial activity \(A_0\) and knowing the mass \(m\), the molar mass \(\mu\), and the Avogadro constant \(N_A\), the half-life can be determined.

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Key Concepts

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

Half-Life Calculation
The half-life of a radioactive isotope is the time taken for half of the nuclei in a sample to decay. It is calculated using the decay constant \(\lambda\) by the formula: \(T_{1/2} = \frac{\text{ln}(2)}{\text{\lambda}}\). The relationship between decay constant and half-life helps understand how quickly a radioactive substance will decrease in activity over time. When the half-life is very long, direct measurement of decay over short periods is impractical, but understanding this relationship allows indirect measurement methods to be utilized. For example, by knowing the initial activity and using the decay constant, one can determine the half-life even for isotopes with minimal observable decay over short times.
Decay Constant
The decay constant \(\text{\lambda}\) represents the probability of a single nucleus decaying per unit time. It is a crucial parameter in understanding radioactive decay. The larger the decay constant, the shorter the half-life of the isotope. The formula \(A = -\frac{dN}{d t} = N_{0}\text{\lambda} e^{-\text{\lambda} t}\) describes how the activity changes over time, where \(N_{0}\) is the initial number of nuclei. At time \(t=0\), the activity is maximized, hence we use this to derive the initial activity as: \(A_{0} = \frac{m N_{A}}{\text{\mu}} \text{\lambda}\). This shows that knowing the decay constant is pivotal in linking the initial activity and the number of nuclei to the half-life.
Initial Activity
Initial activity \(A_{0}\) is the activity of a radioactive sample at the starting point (t = 0). It is given by the formula \(A_{0} = \frac{m}{\text{\mu}} N_{A} \text{\lambda}\), where \(m\) is the mass of the sample, \(\mu\) is the molar mass, \(N_{A}\) is the Avogadro constant, and \(\lambda\) is the decay constant. This formula shows the direct relation between the measurable initial activity and the intrinsic properties of the radioactive isotope. By measuring the initial activity, one can infer details about the radioactive properties of the sample, particularly useful for isotopes with long half-lives where direct observation is impractical.
Number of Nuclei
The number of nuclei \(N_{0}\) in a given mass \(m\) of an isotope can be calculated using the formula \(N_{0} = \frac{m}{\text{\mu}} N_{A}\), where \(\mu\) is the molar mass and \(N_{A}\) is the Avogadro constant. This formula stems from dividing the mass by the molar mass to find the number of moles, then multiplying by the Avogadro constant to convert moles to number of atoms. This calculation is fundamental in understanding the initial quantity of radioactive material in a sample, which is crucial for further calculations of activity and decay.

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Most popular questions from this chapter

How much energy is required to remove one proton from the nucleus of \(^{16}\) O A rough answer to this question is obtained by giving the binding energy per nucleon. A better answer is obtained when we write a reaction that removes a proton from the nucleus. In this case \(^{16} \mathrm{O} \rightarrow_{1}^{1} \mathrm{p}+^{15} \mathrm{N}\) Find the energy required for this reaction to take place. This is the proton separation energy. Get both values and compare them. (The atomic mass of oxygen is \(15.994 \mathrm{u} ;\) that of nitrogen is 15.000 u).

Show that the nuclear density is the same for all nuclei. (Take the masses of the proton and neutron to be the same.)

(a) Show that the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference \(V\) is given by \(\lambda=\frac{h}{\sqrt{2 m e V}}\) (b) Calculate the ratio of the de Broglie wavelength of a proton to that of an alpha particle when both have been accelerated from rest by the same potential difference. (c) Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of \(520 \mathrm{V}\).

singly ionized atoms of neon-20 of velocity \(2.0 \times 10^{5} \mathrm{m} \mathrm{s}^{-1}\) enter the velocity selector of a mass spectrometer, where the magnetic field has the value \(0.15 \mathrm{T}\) (a) What electric field is established in the velocity selector? (b) The ions then enter the region where a second magnetic field of value \(0.50 \mathrm{T}\) deflects them into circular paths. What is the radius of the circular path? (c) If the beam of ions also contains traces of neon-22, what would the detection radius for this isotope be? (Molar masses: neon- \(20=19.992 \mathrm{g} \mathrm{mol}^{-1}\) neon- \(22=21.99 \mathrm{g} \mathrm{mol}^{-1}\).)

The radius of an atomic nucleus is given by the expression $$R=1.2 \times A^{1 / 3} \times 10^{-15} \mathrm{m}$$ (a) Use this expression to find the density of a $$\text { nucleus of iron }\left(_{26}^{56} \mathrm{Fe}\right) \text { in } \mathrm{kg} \mathrm{m}^{-3}$$ (b) How does this density compare with the normal density of iron? (c) If a star with a mass equal to 1.4 times the mass of our sun (solar mass \(=2.0 \times 10^{30}\) kg were to have this density, what should its radius be? (Such stars are formed in the end stage of the evolution of normal stars and are called neutron stars.)

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