Nuclide P decays by emission of ionising radiation to produce nuclide Q. This new nuclide then decays by further emission into nuclide R. The process is shown below, with the appropriate mass numbers (nucleon numbers) and atomic numbers (proton numbers). Which line in the table shows the type of particle emitted at each stage, and the value of X

Answer:

22) F

In the first decay, the atomic number is increased by 1 but the mass number remains the same. This can only be beta decay. In beta decay, a fast moving electron is emitted, the mass number remains the same but the atomic number increases by 1.

In second decay, the atomic number decreases by 2. (This may seem confusing as it says there Z-1. However, you must remember that from Z+1, it went to Z – 1.
Overall change = [Z – 1 – (Z + 1)] = [Z – 1 – Z – 1] = (-2)
This means that this must be alpha decay.

In alpha decay, a helium nucleus is emitted. This means that the mass number decreases by 4 and the atomic number decreases by 2. So X = A – 4