Th234 disintegrates and emits 6β- and 7α-particles to form a stable element. Find the atomic number and mass number of the stable product. Also identify the element.
(IIT-JEE, 2004)
Nuclear reaction is:
90Th234→7 2He4+6−1e0+zXA
Equating atomic number on both sides
90 = 14 + 6 x (-1) + Z
∴ Z = 82
Equating mass number on both sides
238 = 7 x 4 + 6 x 0 + A
∴ A = 206
Thus, element with atomic number 82 and mass number 206 is 82Pb206.