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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 'C' formed (approximately) in the first 10 days are, If initially `10^(20)` nuclei of A is taken

A

`10^(18)`

B

`10^(16)`

C

`10^(17)`

D

`10^(19)`

Text Solution

Verified by Experts

The correct Answer is:
C
Half life of .A. is very less when compared to 10 days. In 10 days total A converted to B.
In 10 days no. of B atom converted to .C. is `(0.693)/(6930) = (2.303)/(10) log " " (10^(20))/(10-^(20) - x)`
Then formed .C. slowly disintegrates. The no. of nuclei of .C. formed approximately is ` = 10^(17)`
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