Question

Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of a and b particles emitted is –

• A. a = 3, b = 3
• B. a = 6, b = 4
• C. a = 6, b = 0
• D. a = 4, b = 6

Answer 1

Answer: (B)

a = 6, b = 4

Answer 2

$\mathrm { n } _ { \alpha } = \frac { 232 - 208 } { 4 } = 6$