Question

In disintegration series of ${}_{92}^{238}U$, the final stable nucleus of X is formed with an overall emission of 8 $\alpha$ and 6 $\beta$ particles. How many neutrons does X nucleus have?

Answer 1

$\alpha$ particles are ${}_2^{4}He$
$\beta$ particles are ${}_{-1}^{0}e$
So,mass number of X $=238-8\times 4=206$
Atomic number of X $=92-8\times 2+6=82$
So, the number of neutrons in X $=206-82=124$