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For sample of ""^(66)Cu, 7/8 of it decays in 15 min. The half-life of sample is: |
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Answer» 5 min `N/N_(0)=(1/2)^(t/T)," here N"=1/8 N_(0) ("AMOUNT LEFT")` `1/8=(1/2)^(t/T) RARR (1/2)^(3)=(1/2)^(t/T) therefore t/T=3` `t=3T rArr 15 =3T therefore T=5"min"` |
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