Question

${}_{90}^{234}Th$ disintegrates to give ${}_{82}^{206}Pb$ as the final product. How many alpha and beta particles are emitted in this process

• A. 6
• B. 8
• C. 9
• D. 2

Answer 1

Answer: (A)

6

Answer 2

$\underset{\text{Parent}}{{}_{90}^{234}Th}\overset{\quad\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}\quad}{\rightarrow}\underset{\text{End product}}{{}_{82}^{206}Pb}$

Decrease in mass = (234 – 206) = 28

Mass of α-particle = 4

So Number of α-particles emitted$= \frac{28}{4} = 7$

Number of beta particles emitted = 2 × No. of α-particles – (At. no. of parent – At. no. of end product)

= 2 × 7 – (90 – 82) = 6