Question

After lossing a number of $\alpha$- and $\beta$- particles ${}_{92}^{238}U$ is changed to ${}_{82}^{206}Pb$. The number of $\alpha$-particles lost in this process is

• A. $10$
• B. $5$
• C. $8$
• D. $32$

Answer 1

The correct option is C $8$

${}_{92}^{238}U{⟶}_{82}^{206}Pb+x_2^4\alpha +y_{-1}^0\beta$

Number of $\alpha$ particles emitted

$=\frac{changeinatomicmass}{4}$

$\frac{238-206}{4}$

$=\frac{32}{4}=8$