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Question-139108




Question Number 139108 by BHOOPENDRA last updated on 22/Apr/21
Answered by physicstutes last updated on 22/Apr/21
λ = 0.1317 year^(−1) , t = 21.04 years, N_0  = 1 g  Using the decay equation:   N = N_0 e^(−λt)         = (1)(e)^((−0.1317)(21.04)) g = 0.006 g
$$\lambda\:=\:\mathrm{0}.\mathrm{1317}\:\mathrm{year}^{−\mathrm{1}} ,\:{t}\:=\:\mathrm{21}.\mathrm{04}\:\mathrm{years},\:{N}_{\mathrm{0}} \:=\:\mathrm{1}\:\mathrm{g} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{decay}\:\mathrm{equation}: \\ $$$$\:{N}\:=\:{N}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\:\:\:\:\:\:=\:\left(\mathrm{1}\right)\left({e}\right)^{\left(−\mathrm{0}.\mathrm{1317}\right)\left(\mathrm{21}.\mathrm{04}\right)} \mathrm{g}\:=\:\mathrm{0}.\mathrm{006}\:\mathrm{g} \\ $$

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