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1 g of .(79)Au^(198) (t(1//2) = 65 hr) d...

`1 g` of `._(79)Au^(198) (t_(1//2) = 65 hr)` decays by `beta`-emission to produce stable `Hg`.
a. Write nuclear reaction for process.
b. How much `Hg` will be present after 260 hr.

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(a) `._(79)^(198)Au to ._(80)^(198)Hg + ._(-1)^(6)e`
(b) No. of half lives in 260 hours `= (260)/(65) = 4`
Amount of gold left after 4 half livess `= ((1)/(2))^(4) = (1)/(16)g`
Amount of gold disintegrated `= 1 - (1)/(16) = (15)/(16) g`
Amount of mercury formed `= (15)/(16) = 0.9375 g`
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