Question

The atomic number and mass number (M) of the nucleide formed when three alpha $ (\alpha ) $ and two beta $ (\beta ) $ particles are emitted from $ {{\,}_{\text{92}}}{{\text{U}}^{\text{238}}}\text{.} $

• A. $ A=87,M=233 $
• B. $ A=86,M=226 $
• C. $ A=88,M=235 $
• D. $ A=88,M=226 $

Answer 1

The Correct Option is D

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\times 3=6 $ unit and atomic weight is decreased by $ 3\times 4=12 $ unit. The atomic number and atomic weight of the new nucleide will be: $ {{\,}_{A}}{{X}^{M}}\xrightarrow{-3\alpha }{{\,}_{A-6}}{{X}^{M-12}} $ After emitting two beta particles the atomic number is increased by $ 1\times 2=2 $ unit. So, the atomic number and atomic weight of new nucleide will be $ {{\,}_{A-6}}{{X}^{M-12}}{{\xrightarrow{-2\beta }}_{A-6+2}}\,{{X}^{M-12}} $ or $ {{\,}_{A-4}}{{X}^{M-12}} $ e, $ {{\,}_{92}}{{U}^{238}}\xrightarrow{-3\alpha }{{\,}_{86}}{{X}^{226}}\xrightarrow{-2\beta }{{\,}_{88}}{{X}^{226}} $ So, the atomic number = 88 and atomic weight = 226