Question

In successive radioactive decay if the decrease in mass number is 32 and
       the decrease in atomic number is 8, how many   - particles and - particles
                 are emitted in the process​

Answer 1

Answer:

Solution : The change in mass is 238 - 206 = 32 unit. It means that 32/4=8α-particles are emitted

Explanation:

PLEASE MARK ME AS A BRIANLIST

Answer 2

α particles emitted = 8

β particles emitted = 8

Number of α particles emitted = $\frac{Change in atomic mass number}{4}$

∴ Number of α particles emitted = $\frac{32}{4}$

∴ Number of α particles emitted = 8

Number of β particles emitted = 2 × number of α particles - change in atomic nubmer

∴ Number of β particles emitted = 2 × 8 - 8

∴ Number of β particles emitted = 16 - 8

∴ Number of β particles emitted = 8

α-decay (alpha decay)

It's a sort of decay in which a nucleus emits alpha particles and converts into a separate nucleus. The alpha particle is the helium nucleus (4He2). Helium has two neutrons and two protons, as we know. As a result, following the emission, the atomic mass of the nucleus that is emitting decreases to 4 and the atomic number decreases to 2.

Beta decay (β - decay)

It is a term used to describe the process of the spontaneous disintegration of a nucleus by the emission of solely electrons or positrons is known as beta decay. There are two forms of beta decay: beta plus decay and beta minus decay.