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6. The radioactive isotope (183 Ta) has a half-life of 5.11 days. A sample containing this isotope has an initial activity atplease note that the answer is not 2.4*10^14. please show your work clearly thank you, i'll upvote

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T = = 5ill days sillx24 los = 122. 64 br. nitially mating a IN nd N = d No 이 da 0.693 TI 4.5*10* = 0.693 x No (122.64 x 60x60N = 1.68319x1094 Nuceli remaining – 10/122.64 14 / Na = 2.063 1.38893 x 10 14 N = no of so nudei decayed b/w 36 her and 70 m.

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