Question

Calculate the values of x and y in the following nuclear reaction.

\[\ce{^227_89Ac -> ^211_82Pb + x[^4_2He]+ y[^0_-1e]}\]

Answer 1

Given: Initial atomic mass A_m = 227

Initial atomic number Z_i = 89

Final atomic mass A_d = 211

Final atomic number = 82

x = number of `""_2^4"He"` i.e. α-particles

y= number of `""_-1^0"He"` i.e. β-particles

`4xxalpha(""_2^4"He")=A_m-A_d`

`alpha=(A_m-A_d)/4`

`alpha = (227-211)/4=16/4=4=x`

`82+2x-1y=89`

`82+2xx4-y=89`

`82+8-y=89`

`y=90-89`

`y=1`

`x=4 andy=1`