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The isotope .(92)^(235)U decays in a num...

The isotope `._(92)^(235)U` decays in a number of steps to an isotope of `._(82)^(207)Pb`. The groups of particle emitted in this process will be:

A

`4 alpha , 7beta`

B

`6 alpha , 4 beta`

C

`7 alpha , 4 beta`

D

`10 alpha, 8 beta`

Text Solution

Verified by Experts

The correct Answer is:
C
`""_(92)U^(235) to ""_(82)Pb^(207) + x_2 He^(4) + y ""_(-1)beta^0`
No. of `alpha`-particles = `(235 -207)/(4) = (28)/(4) = 7alpha`
No. of `beta` particles ` =92-82-2xx7=4beta`
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