Let’s identify the missing isotopes in the following nuclear reactions:
a) 23Na + 1H → ? + n
In this reaction, Sodium-23 (23Na) reacts with a proton (1H) to produce a neutron (n) and an unknown isotope. To find the missing isotope, we need to balance the atomic numbers and mass numbers on both sides of the reaction.
Atomic number of Na = 11, Atomic number of H = 1. Total atomic number on the left side = 11 + 1 = 12.
Mass number of Na = 23, Mass number of H = 1. Total mass number on the left side = 23 + 1 = 24.
Since a neutron (n) is produced, its atomic number is 0 and mass number is 1. Therefore, the missing isotope must have an atomic number of 12 and a mass number of 23 (24 – 1).
The missing isotope is 23Mg (Magnesium-23).
b) 6Li + 1H → ?
In this reaction, Lithium-6 (6Li) reacts with a proton (1H) to produce an unknown isotope. Let’s balance the atomic numbers and mass numbers.
Atomic number of Li = 3, Atomic number of H = 1. Total atomic number on the left side = 3 + 1 = 4.
Mass number of Li = 6, Mass number of H = 1. Total mass number on the left side = 6 + 1 = 7.
The missing isotope must have an atomic number of 4 and a mass number of 7.
The missing isotope is 7Be (Beryllium-7).
c) 15O + 1H → ? + 4He
In this reaction, Oxygen-15 (15O) reacts with a proton (1H) to produce Helium-4 (4He) and an unknown isotope. Let’s balance the atomic numbers and mass numbers.
Atomic number of O = 8, Atomic number of H = 1. Total atomic number on the left side = 8 + 1 = 9.
Mass number of O = 15, Mass number of H = 1. Total mass number on the left side = 15 + 1 = 16.
Helium-4 has an atomic number of 2 and a mass number of 4. Therefore, the missing isotope must have an atomic number of 7 (9 – 2) and a mass number of 12 (16 – 4).
The missing isotope is 12N (Nitrogen-12).