Animated Solution for Physics - Atoms and Nuclei: A nuclear transformation is denoted by $X(n,\alpha )\to {}_3^7\text{Li}$. Which of the following is the nucleus of element $X$?

Imagine you are reading a secret code used by nuclear physicists. The question presents us with a nuclear transformation written in a compact shorthand notation: $X(n,\alpha )\to {}_3^7\text{Li}$.

What does this actually mean? Let's break it down. The general format for this notation is $\text{Target}(\text{Projectile},\text{Emitted Particle})\text{Product}$.

In our specific case: - $X$ is the target nucleus we are trying to identify. - $n$ is the projectile, which is a neutron. - $\alpha$ is the emitted particle, which is an alpha particle. - ${}_3^7\text{Li}$ is the final product nucleus.

Translating this into a standard nuclear reaction equation, we get:

Conservation Laws

I know nuclear equations can sometimes look intimidating, but let's take a breath. They are governed by two incredibly simple and powerful conservation laws. In any low-energy nuclear reaction, two quantities must remain perfectly balanced on both sides of the arrow:

1. Conservation of Mass Number ($A$): The total number of nucleons (protons + neutrons) must be the same before and after the reaction. 2. Conservation of Atomic Number ($Z$): The total electrical charge (number of protons) must be conserved.

Let's represent our unknown nucleus $X$ with its atomic number $Z$ and mass number $A$ as ${}_Z^A\text{X}$. We also need to recall the standard notations for our particles: - A neutron is written as ${}_0^1\text{n}$ because it has a mass of $1$ atomic mass unit and a charge of $0$. - An alpha particle is a helium nucleus, written as ${}_2^4\text{He}$, with a mass of $4$ and a charge of $+2$.

Substituting these into our equation, we get the raw setup:

Now, let's apply our conservation laws to solve for the unknowns.

First, let's balance the mass numbers (the superscripts): On the left side, we have $A$ from our unknown nucleus and $1$ from the neutron. On the right side, we have $7$ from Lithium and $4$ from the alpha particle.

Next, let's balance the atomic numbers (the subscripts): On the left side, we have $Z$ from our unknown nucleus and $0$ from the neutron. On the right side, we have $3$ from Lithium and $2$ from the alpha particle.

We have successfully deduced the properties of our mystery element. It has an atomic number $Z=5$ and a mass number $A=10$.

To identify the element, we simply look at the atomic number. Which element in the periodic table sits at position number $5$? It is Boron (B).

Therefore, our unknown target nucleus is Boron-10, written as ${}_5^{10}\text{B}$. This perfectly matches option (b).