Two radioactive nuclides A and B have half lives of 50 min and 10 min respectively. A fresh sample contains the nuclides of B to be eight time that of A. How much time should elapse so that the number of nuclides of A becomes double of B
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Radioactive decay follows the formula: , where λ = ln(2)/T (T is half-life). For A: λA = ln(2)/50, for B: λB = ln(2)/10. Initially, NB0 = 8NA0. We need time t when NA = 2NB. Substituting into decay laws: NA0e-λAt = 2 * 8NA0e-λBt. Simplifying: e(λB - λA)t = 16. Taking ln: t(ln(2)/10 - ln(2)/50) = ln(16). Solving: t = (4ln(2)) / (ln(2)(0.1 - 0.02)) = 4 / 0.08 = 50 min.
Final Answer: 50