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Four vessels A,B,C and D contain respectively 20g atom (T_(1//2) = 5h) 2g atom (T_(1//2)=1h) 5g atom (T_(1//2) = 2h)and 10g atom (T_(1//2) = 3h) of different radio nuclides in the 12 beginning, the maximum activity would be exhibited by the vessel is |
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Answer» SOLUTION :`R=lambdaN=0.693/T_(1//2)N` Hence, greater `(N/T_(1//2))` value , greater will be the RATE .So A has Maximum ACTIVITY. |
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