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(B) Given, half-life of the radioactive substance, $ T_{1/2} = 20 	ext{ minutes} $.At $ 50\% $ decay, the remaining amount $ N_1 = 50\% $ of $ N_0 =

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$ 40 	ext{ minutes} $

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(B) Given, half-life of the radioactive substance, $ T_{1/2} = 20 	ext{ minutes} $.At $ 50\% $ decay, the remaining amount $ N_1 = 50\% $ of $ N_0 = 0.5 N_0 $. This corresponds to $ 1 $ half-life, so $ t_1 = 20 	ext{ minutes} $.At $ 87.5\% $ decay, the remaining amount $ N_2 = (100 - 87.5)\% $ of $ N_0 = 12.5\% $ of $ N_0 = 0.125 N_0 = (1/8) N_0 = (1/2)^3 N_0 $.This corresponds to $ 3 $ half-lives, so $ t_2 = 3 	imes T_{1/2} = 3 	imes 20 = 60 	ext{ minutes} $.The time taken between $ 50\% $ decay and $ 87.5\% $ decay is $ \Delta t = t_2 - t_1 = 60 - 20 = 40 	ext{ minutes} $.

The half-life of a radioactive substance is $ 20 \text{ minutes} $. The time taken between $ 50\% $ decay and $ 87.5\% $ decay of the substance will be

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