Question

\({ }_{3} \mathrm{Li}^{7}+{ }_{1} p^{1} \longrightarrow X\); Identify \(X\) if reaction is \((p, \alpha)\) type. (a) \(_{4} \mathrm{Be}^{8}\) 1 (b) \({ }_{2} \mathrm{He}^{4}\) (c) \({ }_{0} \gamma^{0}\) (d) none of these \(x\)

Short answer

The product X in the reaction is _{2}He^{4}, which is the alpha particle (helium-4 nucleus). The correct answer is option (b).

Step-by-step solution

Understand the Reaction Type

A (p, α) reaction involves a proton (_{1}p^{1}) entering a nucleus and an alpha particle (_{2}He^{4}) being emitted. The target nucleus here is _{3}Li^{7}. Therefore, we need to determine the product nucleus X after the alpha particle is released.

Determine the New Nucleus

To find X, balance the reaction by conserving the number of protons and neutrons. Given that an alpha particle is emitted, subtract its protons and neutrons from the original lithium nucleus: (_{3}Li^{7})_{protons} + (_{1}p^{1})_{proton} - (_{2}He^{4})_{protons + neutrons} = (_{2}X^{4})_{protons + neutrons}. Therefore, X should have 2 protons and a mass number of 4, which is _{2}He^{4}.

Choose the Correct Answer

The correct answer must have 2 protons and a mass number of 4 to conserve the number of protons and the mass number. The correct answer is (b), which represents an alpha particle or helium-4 nucleus _{2}He^{4}.

Key concepts

Proton-Alpha Reaction

Understanding the proton-alpha reaction is essential for grasping nuclear reaction problems. This type of reaction typically involves a lightweight nucleus like lithium being struck by a proton. The collision results in the emission of an alpha particle, which is essentially a helium-4 nucleus composed of two protons and two neutrons.

The alpha particle is represented by the symbol _{2}^{4}He. Since we're starting with a lithium-7 nucleus _{3}^{7}Li and adding a proton _{1}^{1}p to it, we can expect an alteration in the makeup of the lithium nucleus after the emission of the alpha particle. The final product of this interaction must account for both the particles added and removed, ensuring that both the charge (or atomic number) and the mass number are balanced, abiding by the law of conservation of mass and energy.

Nuclear Equation Balancing

Balancing a nuclear equation is like solving a puzzle, where we must make sure that the number of protons and neutrons is conserved before and after the reaction. When an alpha particle (_{2}^{4}He) is released, we subtract its mass and atomic numbers from the original nucleus.

For the given exercise, the mass number of lithium-7 and a proton add up to 8 (7+1=8), but since we emit an alpha particle with a mass number of 4, our new nucleus must have a mass number of 4 (8-4=4). Similarly, the resultant nucleus will have an atomic number decreased by 2 (3+1-2=2), as two protons have been emitted with the alpha particle. The process of balancing involves ensuring that we match these numbers to find the correct identity of the final nucleus, X.

Alpha Particle Emission

Alpha particle emission is a common type of radioactive decay and is also an integral part of certain nuclear reactions, such as the proton-alpha reaction. An alpha particle, denoted by _{2}^{4}He, is a tightly bound group of two protons and two neutrons. Its emission from a nucleus results in a new element that is two atomic numbers lower and has a mass number that is four units smaller.

In our example, after the emission of an alpha particle from the lithium-7 nucleus which has interacted with a proton, the lithium nucleus loses those two protons and two neutrons. This transforms it into a lighter element with fewer protons, hence the atomic number changes from 3 to 2. The material that remains is a different isotope of helium, with a mass number reduced by 4 units to reflect the loss of the alpha particle. The understanding of alpha emission helps us solve nuclear equation problems and predict the behavior of radioisotopes.