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Chapter 43: Problem 45

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

A radioactive isotope has a half-life of 43.0 min. At \(t\) = 0 its activity is 0.376 Ci. What is its activity at \(t\) = 2.00 h?

Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of 100 g, which includes \(9.4 \mu Ci\) of \(^{59}Fe\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

The unstable isotope \(^4$$^0\)K is used for dating rock samples. Its half-life is 1.28 \(\times\) 10\(^9\) y. (a) How many decays occur per second in a sample containing 1.63 \(\times\) 10\(^-$$^6\) g of \(^4$$^0\)K? (b) What is the activity of the sample in curies?

Consider the nuclear reaction \(^{28}_{14}Si\) + \(\gamma\) \(\rightarrow\) \(^{24}_{12}Mg\) + X where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \(^{28}_{14}Si\) atom is 27.976927 u, and the mass of a \(^{24}_{12}Mg\) atom is 23.985042 u.

As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 400 \(\mu\)Ci of \(^1$$^3$$^1\)Ba, which has a half-life of 12 days. (a) What mass of \(^1$$^3$$^1\)Ba was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00 \(\mu\)Ci. How long will the lab have to be closed?
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