The atomic number (A) and mass number (M) of the nuclide formed when three alpha `(alpha)` and 2 beta `(beta)` particles are emitted from `""_(92)^(238)U` is

Options

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

Answer

D

Explanation

`""_(92)U^(238) to 3""_(2)He^(4)+ 2""_(-1)e^(0) + ""_(A)X^(M)`
`92=3 xx 2+ 2xx(-1)+ A`
`A =92 -6 + 2=88`
`238 = 3 xx 4 + 2 xx0 + M`
M=238 -12 =226