/*! 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 11 Neutron capture in sodium occurs... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 11

Neutron capture in sodium occurs with a cross section of \(0.53 \mathrm{b}\) and leads to radioactive \({ }^{24} \mathrm{Na}\left(t_{1 / 2}=15 \mathrm{h}\right) .\) What is the activity that results when \(1.0 \mu g\) of \(\mathrm{Na}\) is placed in a neutron flux of \(2.5 \times 10^{13}\) neutrons \(/ \mathrm{cm}^{2} / \mathrm{s}\) for \(4.0 \mathrm{h} ?\)

Short Answer

Expert verified
63840 Bq

Step by step solution

01

- Calculate the number of sodium atoms

First, we need to find the number of sodium atoms in 1.0 \mu g of sodium. The molar mass of sodium (Na) is 23 g/mol. Therefore, the number of sodium atoms is given by \[ N = \frac{1.0 \times 10^{-6} \text{g}}{23 \text{g/mol}} \times 6.022 \times 10^{23} \text{atoms/mol} \]Substitute the values to get \[ N = \frac{1.0 \times 10^{-6}}{23} \times 6.022 \times 10^{23} \approx 2.618 \times 10^{16} \text{atoms} \]
02

- Calculate the number of neutrons interacting with sodium atoms

The neutron flux is given as \[ \Phi = 2.5 \times 10^{13} \text{neutrons}/\text{cm}^2/\text{s} \].The interaction time is 4.0 hours. Convert this time to seconds: \[ t = 4.0 \text{h} \times 3600 \text{s/h} = 14400 \text{s} \].So, the total neutron flux over the interaction time is \[ \Phi_t = 2.5 \times 10^{13} \times 14400 = 3.6 \times 10^{17} \text{neutrons}/\text{cm}^2 \]
03

- Calculate cross-sectional area for neutron capture

The cross section (\sigma) for neutron capture is given as 0.53 b (barn). Convert this value to cm\(^2\): \[ \sigma = 0.53 \text{b} \times 10^{-24} \text{cm}^2/\text{b} = 0.53 \times 10^{-24} \text{cm}^2 \]
04

- Calculate the reaction rate

The reaction rate (R) can be calculated as: \[ R = N \times \Phi_t \times \sigma \].Substitute in the values: \[ R = 2.618 \times 10^{16} \times 3.6 \times 10^{17} \times 0.53 \times 10^{-24} \]Simplify the expression: \[ R = 2.618 \times 3.6 \times 0.53 \times 10^9 = 4.98 \times 10^9 \text{captures} \]
05

- Calculate the activity

Activity (A) is given by: \[ A = R \times \lambda \]where \lambda is the decay constant. The decay constant is related to the half-life by: \[ \lambda = \frac{\ln 2}{t_{1/2}} \].The half-life \( t_{1/2} \) of \( {^{24} \text{Na}} \) is 15 hours. Converting to seconds: \[ t_{1/2} = 15 \times 3600 = 54000 \text{s} \]\[ \lambda = \frac{\ln 2}{54000} \approx 1.283 \times 10^{-5} \text{s}^{-1} \].Therefore, the activity is: \[ A = 4.98 \times 10^9 \times 1.283 \times 10^{-5} \approx 63840 \text{Bq} \]

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Key Concepts

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

Neutron Flux
Neutron flux is a measure of the intensity of neutron radiation at a specific location. It indicates the number of neutrons passing through a unit area per unit time. This is crucial in nuclear reactions as it impacts the rate of reactions. For example, in nuclear reactors, high neutron flux facilitates sustaining the chain reactions. In our problem, the neutron flux is given as \(\text{2.5 x 10}^{13} \text{neutrons}/\text{cm}^2/\text{s}\). Over the interaction time of 4 hours, this accumulates to \(\text{3.6 x 10}^{17} \text{neutrons}/\text{cm}^2\).
Radioactive Decay
Radioactive decay refers to the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur in various forms such as alpha, beta, and gamma decay. In the given exercise, sodium-24 \( {^{24} \text{Na}} \) becomes radioactive upon neutron capture, subsequently undergoing decay. The activity, measured in becquerels (Bq), represents the rate at which these radioactive atoms disintegrate. Understanding this decay process helps account for how rapidly the material's radioactivity diminishes over time.
Half-Life
The half-life of a radioactive isotope is the time required for half the nuclei in a sample to undergo decay. It is a crucial factor in determining the longevity and intensity of the radioactivity of a substance. In this problem, the half-life of \( {^{24} \text{Na}} \) is given as 15 hours. This helps in calculating the decay constant, \(\text{位}\), using the formula \(\text{位} = \frac{\text{ln}2}{t_{1/2}}\). In this case, \(\text{位}\) is approximately \(\text{1.283 x 10}^{-5} \text{s}^{-1}\). This constant plays a vital role in determining the activity.
Neutron Cross Section
The neutron cross section is a measure of the probability of a neutron interaction, such as scattering or absorption, with a target nucleus. It's usually measured in barns \( \text{b} \), where 1 barn equals \( \text{$10}^{-24} \text{cm}^2 \). For neutron capture in sodium, the cross section is \(\text{0.53 b}\), equivalent to \( \text{0.53 x 10}^{-24} \text{cm}^2 \). This value influences how many neutrons will interact with sodium atoms, which is critical to calculate the reaction rate in nuclear physics.

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Most popular questions from this chapter

A 2.0 -mg sample of copper \(\left(69 \%^{63} \mathrm{Cu}, 31 \%^{65} \mathrm{Cu}\right)\) is placed in a reactor where it is exposed to a neutron flux of \(5.0 \times 10^{12}\) neutrons \(/ \mathrm{cm}^{2} / \mathrm{s}\). After \(10.0 \mathrm{min}\) the resulting activities are \(72 \mu\) Ci of \({ }^{64}\) Cu \(\left(t_{1 / 2}=12.7 \mathrm{h}\right)\) and \(1.30 \mathrm{mCi}\) of \({ }^{66} \mathrm{Cu}\left(t_{1 / 2}=5.1 \mathrm{min}\right) .\) Find the cross sections of \({ }^{63} \mathrm{Cu}\) and \({ }^{65} \mathrm{Cu}\).

The radioisotope \({ }^{15} \mathrm{O}\left(t_{1 / 2}=122 \mathrm{s}\right)\) is used to measure respiratory function. Patients inhale the gas, which is made by irradiating nitrogen gas with deuterons \(\left({ }^{2} \mathrm{H}\right)\). Consider a cubical cell measuring \(1.24 \mathrm{cm}\) on each edge, which holds nitrogen gas at a pressure of 2.25 atm and a temperature of \(293 \mathrm{K}\). One face of the cube is uniformly irradiated with a deuteron beam having a current of 2.05 A. At the chosen deuteron energy, the reaction cross section is 0.21 b. \((a)\) At what rate is \({ }^{15}\) O produced in the cell? (b) After an irradiation lasting for \(60.0 \mathrm{s}\), what is the activity of \({ }^{15} \mathrm{O}\) in the cell?

A beam of alpha particles is incident on a target of \({ }^{63} \mathrm{Cu}\), resulting in the reaction \(\alpha+{ }^{63} \mathrm{Cu} \rightarrow{ }^{66} \mathrm{Ga}+\mathrm{n}\). Assume the cross section for the particular alpha energy to be \(1.25 \mathrm{b}\) The target is in the form of a foil of \(2.5 \mu \mathrm{m}\) thickness. The beam has a circular cross section of diameter \(0.50 \mathrm{cm}\) and a current of \(7.5 \mu\) A. Find the rate of neutron emission.

In the reaction \(n+{ }^{7} B e \rightarrow p+{ }^{7} L i\) with \(4.500-M e V\) neutrons incident on Be at rest, the Li are observed moving in the forward direction (the same direction as the incident neutrons) with momentum 176.1 MeV/c. \((a)\) Find the momentum and energy of the outgoing protons. (b) Show how these results could be used to determine the mass of 'Be if the masses of \(\mathrm{p}, \mathrm{n},\) and \({ }^{7} \mathrm{Li}\) are known.

List five nuclear reactions, consisting of a light, stable projectile nucleus (mass 4 or less) incident on a heavy, stable target nucleus that can produce the radioactive nucleus \(^{56}\) Co.
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