Question

How do you write a balanced nuclear equation for the alpha decay of radioactive isotopes? Write a balanced nuclear equation for the alpha decay of each of the following radioactive isotopes.
A) Curium- \[243\]
B) Th- Atomic Mass Number: $232$ and Atomic Number: $90$
C) No- Atomic Mass Number: $251$ and Atomic Number: $102$
D) Radon- $220$

Answer 1

Hint: Alpha decay is a process where an element disintegrates or splits to give a different element and an alpha particle. The alpha particle is a helium atom. The alpha particle or the helium atom is represented as \[{}_2^4He\] where $4$ is the atomic mass number and $2$ is the atomic number.

Complete step-by-step answer:
Alpha decay is the removal or decay of an alpha particle from an element. This alpha particle is a helium atom. In order to write an equation for the alpha decay of a radioactive isotope, we have to first know the atomic mass number and atomic number of both the element and the alpha particle.

Let us take an element X with an atomic mass number of A and atomic number of Z. After alpha decay, the atomic number of the resultant element will be the difference of the atomic number of element X and the atomic number of Helium atom. We can write the same for atomic mass numbers.

Let us say that the resultant element is Y, then we can write the equation for alpha decay of X as follows.
${}_Z^AX \to {}_{Z - 2}^{A - 4}Y + {}_2^4He$

A) The atomic mass number of Curium is $243$ and the atomic number is $96$
Substituting these values in the above equation, we get ${}_{96}^{243}Cm \to {}_{94}^{239}Pu + {}_2^4He$
Here, Pu is Plutonium

B) Similarly for Th, we can write the equation as ${}_{90}^{232}Th \to {}_{88}^{228}Ra + {}_2^4He$
Here, Ra is Radium.

C) For No, we can write the equation for its alpha decay as ${}_{102}^{251}No \to {}_{100}^{247}Fm + {}_2^4He$
Here, Fm is Fermium.

D) Atomic number of Radon is $86$ and atomic mass number is $220$
We can write the equation for alpha decay of Radon as follows.
${}_{86}^{220}Rn \to {}_{84}^{216}Po + {}_2^4He$
Here, Po is Polonium.

Note: It is to be noted that when a reaction for alpha decay of an element is to be written, one should know the element that is being produced after the alpha decay. This is done by knowing the atomic number and mass number of different elements