1.

The isotope ""_(92)^(235) U decays in a number of steps to an isotope of ""_(87)^(207)Pb. The no of particles emitted in this process will be :

Answer»

`4 ALPHA`
`6 beta`
`7 alpha`
`4 beta`

Solution :`""_(92)U^(235) to ""_(82) U^(2-7) + a (""_(2) alpha^(4)) + b(""_(-1) beta^(0))`
`a = (235- 207)/(4) = 7 alpha , 92- 82 = 2 (7) - b , 10 = 14- b IMPLIES b = 4 beta`


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