Question

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

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

Step-by-step solution

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 \[ ^{A}_{Z}X \].

Apply Conservation of Atomic Number (Protons)

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

Apply Conservation of Mass Number (Nucleons)

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

Determine the Nuclide X

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

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**:\[ 10.012938 + 1.008665 = 11.021603 \text{ u} \]\[ Q = (11.018606 - 11.021603) \times 931.5 = -2.79379 \text{ MeV} \]

Conclude Energy Analysis

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

Key concepts

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.