Question

Complete each of the following nuclear equations by supplying the missing particle. a. $?\to {}_2^4\mathrm{He}+{}_{83}^{192}\mathrm{Bi}$ b. $?\to {}_2^4\mathrm{He}+{}_{82}^{204}\mathrm{Pb}$ $\mathrm{c}.?\to {}_2^4\mathrm{He}+{}_{84}^{206}\mathrm{Po}$

Short answer

The missing particles for the given nuclear equations are: a. ${}_{81}^{188}\mathrm{Tl}\to {}_2^4\mathrm{He}+{}_{83}^{192}\mathrm{Bi}$ b. ${}_{80}^{200}\mathrm{Hg}\to {}_2^4\mathrm{He}+{}_{82}^{204}\mathrm{Pb}$ c. ${}_{82}^{202}\mathrm{Pb}\to {}_2^4\mathrm{He}+{}_{84}^{206}\mathrm{Po}$

Step-by-step solution

Equation (a)

For the first equation: $?\to {}_2^4\mathrm{He}+{}_{83}^{192}\mathrm{Bi}$ To find the missing particle, we can conserve the atomic and mass numbers. $Z_X+2=83\to Z_X=81$ $A_X+4=192\to A_X=188$ Now, we have the missing particle's atomic number and mass number as 81 and 188. Thus, the particle is: ${}_{81}^{188}\mathrm{Tl}\to {}_2^4\mathrm{He}+{}_{83}^{192}\mathrm{Bi}$

Equation (b)

For the second equation: $?\to {}_2^4\mathrm{He}+{}_{82}^{204}\mathrm{Pb}$ To find the missing particle, we can conserve the atomic and mass numbers again. $Z_X+2=82\to Z_X=80$ $A_X+4=204\to A_X=200$ Now, we have the missing particle's atomic number and mass number as 80 and 200. Thus, the particle is: ${}_{80}^{200}\mathrm{Hg}\to {}_2^4\mathrm{He}+{}_{82}^{204}\mathrm{Pb}$

Equation (c)

For the third equation: $?\to {}_2^4\mathrm{He}+{}_{84}^{206}\mathrm{Po}$ To find the missing particle, we can conserve the atomic and mass numbers. $Z_X+2=84\to Z_X=82$ $A_X+4=206\to A_X=202$ Now, we have the missing particle's atomic number and mass number as 82 and 202. Thus, the particle is: ${}_{82}^{202}\mathrm{Pb}\to {}_2^4\mathrm{He}+{}_{84}^{206}\mathrm{Po}$

Key concepts

Atomic Number

The atomic number of an element is a fundamental property that defines the type of atom it is. It represents the number of protons present in the nucleus of an atom. Protons are positively charged particles, and the atomic number gives the element its identity. In the periodic table, elements are arranged in order of increasing atomic number.

For example:

• The atomic number of hydrogen is 1 because it has one proton.
• Helium, with two protons, has an atomic number of 2.

Understanding atomic numbers is crucial when writing or solving nuclear equations, as seen in the original exercise. In these equations, the atomic number helps balance nuclear reactions by ensuring the conservation of charge.

An important part of nuclear equations is keeping track of changes in atomic numbers. During decay processes such as alpha decay, which will be discussed later, the atomic number decreases, influencing the identity of the resulting element.

Mass Number

The mass number of an atom is the total count of protons and neutrons in the nucleus. It is not to be confused with the atomic number, which only counts protons. Neutrons add to the mass of an atom but do not affect its charge. Because electrons, which orbit the nucleus, have negligible mass, they are not considered in the mass number.

Let's break it down:

• Mass number = Number of protons + Number of neutrons.
• It's represented as the superscript number on the left side of an element's symbol, like in ${}_{83}^{192}\mathrm{Bi}$, where 192 is the mass number.

Mass numbers are vital for identifying isotopes, which are atoms of the same element but with different numbers of neutrons. These isotopes have different properties, especially in nuclear reactions such as alpha decay. In nuclear equations, such as those in the original exercise, the conservation of mass number helps to complete equations by balancing the total number of protons and neutrons before and after the reaction.

Alpha Decay

Alpha decay is a common type of radioactive decay where an unstable atomic nucleus loses energy by emitting an alpha particle. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons. This particle is represented by ${}_2^4\mathrm{He}$ in nuclear equations.

In alpha decay:

• The mass number of the original atom decreases by 4 (since two neutrons and two protons are lost).
• The atomic number decreases by 2 (two protons are emitted with the alpha particle).

This process results in the formation of a new element, two steps back in the periodic table, due to the loss of protons. This decay can be seen in the original exercise, where each reaction involves an atom undergoing alpha decay, losing its helium nucleus, and transforming into a different element.

Alpha decay is critical in nuclear physics because it changes one element into another, helping scientists understand the forces and changes occurring in unstable nuclei. Understanding alpha decay is essential to balance nuclear equations effectively, ensuring mass and atomic numbers are preserved across the transaction.