Question

What are the respective number of α and β-particles emitted in the following radioactive decay?
(A) 8 and 8
(B) 8 and 6
(C) 6 and 8
(D) 6 and 6

Answer 1

Hint: The alpha particle is a doubly ionized helium atom, beta is a fast moving electron and gamma is photon. Using this and the changes made by the individual emissions the final A and Z can be calculated. Since the product is given to us already we use the same values to calculate the number of particles emitted.

Complete step-by-step answer
The radioactivity phenomenon can be understood as the emission of radiations by heavy elements spontaneously. They are reoffered as radioactive elements.
There are 3 types if decay:
α-decay: N and Z values decrease by two and A increases by four.
β-decay :N decreases by one, Z increases by one and A does not change.
γ-decay : no change occurs to the parent atom after the decay.
$${}^{200}{X}_{90}\to {}^{168}{Y}_{80}$$
here, the change in A is done only by alpha, it is reduced to 168 from 200.
$200-168=32$
Each alpha particle decreases A by four. So,
${\displaystyle \frac{32}{4}}=8$
Hence, 8 alpha particles are emitted. For one alpha particle Z is reduced by two so for eight alpha particles Z should be reduced by 16 which is 74. But the product has 6 extra.
Each beta particle increases Z by 1 so 6 particles have to be emitted to get Z as 80.

Therefore, 8 alpha and 6 beta particles are emitted and the correct option is B.

Note: Alpha decay occurs principally with nuclei that are too large to be stable. Beta decay occurs when the neutral atomic mass of the original atom is larger than that of the final atom. It is the emission of photons in gamma decay.