Question

Find out the total number of α- and β-particles emitted in the disintegration of $90Th^{232}$ to $82Pb^{208}$

• A. 6 and 4
• B. 8 and 4
• C. 9 and 6
• D. 2 and 4

Answer 1

Answer: (A)

6 and 4

Answer 2

Change in at. wt. = 232 – 208 = 24 amu (at. wt. unit) Now since in one α-particle emission, at. wt. is decreased by 4 amu, the number of α-emissions for 24 amu = 24/4 = 6

Atomic number after 6α-emissions = 90 – 12 = 78$(\because\alpha =_{2}He^{4})$

Increase in atomic number from 78 to the given 82 = 82 – 78

= 4 $(\because\beta\text{-particle} =_{- 1}e^{0})$

∴ No. of β-particle emissions = 4