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The half -life period of .(84)Po^(210) i...

The half -life period of `._(84)Po^(210)` is 140 days.
In how many days `1 g` of this isotope is reduced to `0.25 g`?

Text Solution

Verified by Experts

Original quantity of the isotope `(N_(0)) = 1 g`
Final quantity of the isotope `N = 0.25 g`
We know that, `N = ((1)/(2))^(n) N_(0)`
So, `(1)/(4) = ((1)/(2))^(n) xx 1`
or `((1)/(2))^(2) = ((1)/(2))^(2)`
or `n = 2`
Time taken `T = n xx t_(1//2) = 2 xx 140 = 280` days
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