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Ac89^231 gives Pb82^207 after emission...

`Ac_89^231 gives Pb_82^207` after emission of some `alpha and beta`-particles. The number of such  and -particles are respectively:

A

5, 6

B

6, 5

C

7, 5

D

5, 7

Text Solution

Verified by Experts

The correct Answer is:
B
Number of `alpha-"particles " = (231 -207)/(4) = 6`
Number of `beta-"particles" = 2xx 6 - 89 + 82`
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