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Assuming that .^(226)Ra (t(1//2) = 1.6 x...

Assuming that `.^(226)Ra (t_(1//2) = 1.6 xx 10^(3) yrs)` is in secular equilibrium with `.^(238)U (t_(1//2) = 4.5 xx 10^(9) yrs)` in a certain mineral how many grams of radium will present in for every gram of `.^(238)U` in the mineral?

A

`3.7 xx 10^(-7)`

B

`3.4 xx 10^(7)`

C

`3.4 xx 10^(-7)`

D

`3.7 xx 10^(7)`

Text Solution

Verified by Experts

The correct Answer is:
c
`(N_(1) .^(226)Ra)/(N_(2) .^(238)U) = (t_(1//2) .^(226)Ra)/(t_(1//2) .^(238)U)`
`(w // 226)/(1 // 238) = (1.6 xx 10^(3))/(4.5 xx 10^(9))` , `w = 3.4 xx 10^(-7) g`
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