Question

The number of alpha and $\beta$ particles emitted when ${}_{92}^{238}U$ is finally converted into ${}_{82}^{206}Pb$ is:

• A. 2,1
• B. 4,2
• C. 8,6
• D. 6,4

Answer 1

The correct option is C 8,6

Let’s say x alpha particles and y beta particles are emitted.
The equation is
${}_{92}^{238}U{\to }_{82}^{206}Pb+x_{2}H{e}^4+y_{-1}{e}^0$
Let’s equate the sum of atomic numbers and mass numbers on both sides.
For mass numbers
238=206+4x ….(i)
For atomic numbers
92 =82+2x-y ……(ii)
Solving both equations, we get x= 8, y = 6