Question

The atomic mass of helium is 232 and its atomic number is 90. During the course of its radioactive disintegration `6 alpha` and `4 beta`-particles are emitted. What is the atomic and atomic number of the final atoms?

Answer 1

Decrease in mass due to emission of `6 alpha` particles `= 6 xx 4 = 24`
So, Atomic mass of the product atom = (232 - 24) = 208)
No of `beta` particles emitted = 2 `xx` No . of `alpha` particle
`- (Z_("Thorium") - Z_("Final atom"))`
`4 = 2 xx 6 - (90 - Z_("Final atom"))`
or `Z_("Final atom") = 82`