Question
Physics Question on Nuclear physics
The atomic number and mass number (M) of the nucleide formed when three alpha (α) and two beta (β) particles are emitted from 92U238.
A=87,M=233
A=86,M=226
A=88,M=235
A=88,M=226
A=88,M=226
Solution
When one alpha particle is emitted then atomic number is decreased by 2 unit and atomic weight is decreased by 4 unit; and when one beta particle is emitted then atomic number is increased by one unit and atomic weight remains the same. So, when three alpha particles emits then atomic number is decreased by 2×3=6 unit and atomic weight is decreased by 3×4=12 unit. The atomic number and atomic weight of the new nucleide will be: AXM−3αA−6XM−12 After emitting two beta particles the atomic number is increased by 1×2=2 unit. So, the atomic number and atomic weight of new nucleide will be A−6XM−12−2βA−6+2XM−12 or A−4XM−12 e, 92U238−3α86X226−2β88X226 So, the atomic number = 88 and atomic weight = 226