Question

${}_{90}^{234}\text{Th}$ disintegrates to give ${}_{82}^{206}\text{Pb}$ as the final product. Number of $\alpha$ and $\beta$-particles are emitted out during this process is :

• A. $[5\alpha ,7\beta ]$
• B. $[7\alpha ,6\beta ]$
• C. $[7\alpha ,5\beta ]$
• D. none of the above

Answer 1

The correct option is B $[7\alpha ,6\beta ]$

$\alpha -$ particle - ${}_2^{4}He$

$\beta -$ particle- ${}_{-1}^{0}e$

${}_{90}^{234}Th{\to }_{82}^{206}Pb+a_2^{4}He+b_{-1}^{0}e$

$90+2a-b=82$

$234-4a=206$

$a=7,b=6$