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A radioactive substance 'A' converts to ...

A radioactive substance 'A' converts to stable nuclei D by following series of reaction.
`A to B to C to D`, Given, `t_(1//2)` for 'A' = 0.0693 days
`t_(1//2)` for 'B' = 6930 days `t_(1//2)` for 'C' = 6.93 days
Number of nuclei of 'D' present after 6930 days are, if initially `10^(20)` nuclei of A is taken

A

`10^(10)`

B

`(1)/(2) xx 10^(20)`

C

`(1)/(2) xx 10^(17)`

D

`10^(9)`

Text Solution

Verified by Experts

The correct Answer is:
B
Half life of A, and C are very less. So `B to D` is 6930 year, No. of nuclei .D. formed is `(10^(20))/(2)`.
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