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Consider the nuclear reaction \(^{4}_{2}He\) + \(^{7}_{3}Li\) \(\rightarrow\) X + \(^{1}_{0}n\) where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Is energy absorbed or liberated? How much?

Short Answer

Expert verified
X is \(^ {10}_{5} B\). Energy is absorbed: 2.79 MeV.

Step by step solution

01

Identify the Components

In this nuclear reaction, \[ ^{4}_{2}He \] is an alpha particle. The target is \[ ^{7}_{3}Li \], and the product includes a neutron \[ ^{1}_{0}n \]. The unknown nuclide X can be represented as \[ ^{A}_{Z}X \].
02

Apply Conservation of Atomic Number (Protons)

By conserving the atomic number (Z), we have:\[ 2 + 3 = Z + 0 \]Therefore,\[ Z = 5 \].
03

Apply Conservation of Mass Number (Nucleons)

By conserving the mass number (A), we have:\[ 4 + 7 = A + 1 \]Therefore,\[ A = 10 \].
04

Determine the Nuclide X

With \[ Z = 5 \] and \[ A = 10 \], the unknown nuclide X is \[ ^{10}_{5}B \], which is boron.
05

Energy Consideration - Calculate the Energy Q-value

First, find the masses of the particles involved:- Mass of \[ ^{4}_{2}He \] = 4.002602 u- Mass of \[ ^{7}_{3}Li \] = 7.016004 u- Mass of \[ ^{10}_{5}B \] = 10.012938 u- Mass of \[ ^{1}_{0}n \] = 1.008665 uCalculate the Q-value using:\[ Q = (\text{Initial Masses} - \text{Final Masses}) \times 931.5 \text{ MeV/c}^2 \]**Initial Masses**:\[ 4.002602 + 7.016004 = 11.018606 \text{ u} \]**Final Masses**:\[ 10.012938 + 1.008665 = 11.021603 \text{ u} \]\[ Q = (11.018606 - 11.021603) \times 931.5 = -2.79379 \text{ MeV} \]
06

Conclude Energy Analysis

Since the Q-value is negative, the reaction absorbs energy (endothermic reaction).

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Key Concepts

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

Alpha Particle
An alpha particle is a type of nuclear particle consisting of two protons and two neutrons. This gives it the characteristics of a helium nucleus. Because alpha particles are relatively heavy and positively charged, they do not travel very far in matter and can be stopped by a piece of paper or skin.

Here are some key takeaways:
  • They have a mass number of 4 and an atomic number of 2, corresponding to the symbol \(^4_2He\).
  • Alpha particles are commonly emitted in radioactive decay processes.
  • They interact strongly with matter, making them useful for smoke detectors but dangerous if ingested.
Understanding the role of alpha particles in nuclear reactions like the one discussed here is crucial. In the given reaction, the alpha particle collides with another nucleus, triggering further reactions and transformations.
Conservation of Atomic Number
The conservation of atomic number is a fundamental principle in nuclear reactions. It states that the total number of protons (atomic number) before the reaction must equal the total number of protons after the reaction.

In other words, nuclear reactions cannot create or destroy the charge. They can only rearrange it. This principle is vital for balancing nuclear reaction equations.

For the given reaction:
  • The alpha particle contributes 2 protons.
  • The lithium target has 3 protons.
  • Together they total 5 protons before the reaction.
On the right side of the equation, any unknown nuclide produced must also have a total of 5 protons. This shows us that the resulting element has an atomic number (Z) of 5, which corresponds to boron.
Energy Q-value
The energy Q-value in nuclear reactions tells us whether energy is absorbed or released during the process. It is calculated based on the difference between the total masses of the reactants and the products. If the Q-value is positive, the reaction releases energy (exothermic).

If it is negative, the reaction absorbs energy (endothermic). In this reaction, the Q-value calculation showed a negative result:
  • Initial mass was 11.018606 u, and final mass was 11.021603 u.
The slight increase in mass indicates energy absorption, calculated as:
  • \[ Q = (11.018606 - 11.021603) \times 931.5 \text{ MeV/c}^2 = -2.79379 \text{ MeV} \]
Endothermic reactions like this one require energy input for the reaction to occur. So, in practical applications, additional energy would need to be supplied for this particular reaction to proceed.

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Most popular questions from this chapter

Which type of radioactive decay produces \(^{131}\)I from \(^{131}Te\)? (a) Alpha decay; (b) \(\beta ^-\) decay; (c)\(\beta ^+\) decay; (d) gamma decay.

\(\textbf{Radioactive Tracers}\). Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. One such tracer is \(^1$$^3$$^1\)I, a \(\beta$$^-\) emitter with a half-life of 8.0 d. Suppose a scientist introduces a sample with an activity of 325 Bq and watches it spread to the organs. (a) Assuming that all of the sample went to the thyroid gland, what will be the decay rate in that gland 24 d (about 3 \\(\frac{1}{2}\\) weeks) later? (b) If the decay rate in the thyroid 24 d later is measured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What isotope remains after the I-131 decays?

Consider the fusion reaction \(^{2}_{1}H\) + \(^{2}_{1}H \rightarrow ^{3}_{2}He + ^{1}_{0}n\). (a) Estimate the barrier energy by calculating the repulsive electrostatic potential energy of the two \(^{2}_{1}H\) nuclei when they touch. (b) Compute the energy liberated in this reaction in MeV and in joules. (c) Compute the energy liberated \(per\) \(mole\) of deuterium, remembering that the gas is diatomic, and compare with the heat of combustion of hydrogen, about \(2.9 \times 10^{5} J/mol\).

\(\textbf{Comparison of Energy Released per Gram of Fuel.}\) (a) When gasoline is burned, it releases 1.3 \(\times\) 10\(^8\) J of energy per gallon (3.788 L). Given that the density of gasoline is 737 kg/m\(^3\), express the quantity of energy released in J/g of fuel. (b) During fission, when a neutron is absorbed by a \(^2$$^3$$^5\)U nucleus, about 200 MeV of energy is released for each nucleus that undergoes fission. Express this quantity in J/g of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one \(^4\)He nucleus with two leftover protons and the liberation of 26.7 MeV of energy. The fuel is the six protons. Express the energy produced here in units of J/g of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other. (d) Our sun produces energy at a measured rate of 3.86 \(\times\) 10\(^2$$^6\) W. If its mass of 1.99 \(\times\) 10\(^3$$^0\) kg were all gasoline, how long could it last before consuming all its fuel? (\(Historical\) \(note\): Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)

At the beginning of Section 43.7 the equation of a fission process is given in which \(^2$$^3$$^5\)U is struck by a neutron and undergoes fission to produce \(^1$$^4$$^4\)Ba, \(^8$$^9\)Kr, and three neutrons. The measured masses of these isotopes are 235.043930 u (\(^2$$^3$$^5\)U), 143.922953 u (\(^1$$^4$$^4\)Ba), 88.917631 u (\(^8$$^9\)Kr), and 1.0086649 u (neutron). (a) Calculate the energy (in MeV) released by each fission reaction. (b) Calculate the energy released per gram of \(^2$$^3$$^5\)U, in MeV/g.

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