A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific test site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Choose the closest.

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0.50 g That should be the correct answer :)


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Answer : The correct answer for amount of radioisotope remain in 2030 is 0.619 g .Radioactive Decay is emission of radiations ( in form of alpha , beta particle etc ) by unstable atom . Radioactive decay is FIRST ORDER reaction . So , the equation of first order can be used to find decay constant , amount of radioisotopes or half life . The equation for radioactive decay is given as : Where : N = amount of radioisotope at time t N₀ = amount of radioisotope initially present k = decay constant t = time Half life :It is time when amount of radioisotope decrease to 50 % of its original amount . Half life and decay constant can be related : Following are the steps can be used to determine amount of radioisotope (N) :1) To find decay constant :Given : = 28 yrsDecay constant can be calculated using half life by plugging value in half life formula :On multiplying both side by k On dividing both side by 28 yrs k = 0.02475 yrs⁻¹2) To find amount of radioisotope (N):Given : Amount of radioisotope originally present = 3.5 g Time = 2030 - 1960 = 70 yrs decay constant = 0.02475 yrs⁻¹ Amount of radioisotope (N) = ?Plugging these values in the formula as: can be converted using the formula ( ) ln N - ln (3.5 ) = - 1.7325 (ln 3.5 = 1.253 ) ln N -1.253 = -1.7325 Adding both side 1.253 ln N -1.253 + 1.253 = -1.7325 + 1.253 ln N = -0.4795Taking anti ln of -0.4795 N = 0.619 g Hence amount of radioisotope remained in 2030 is 0.619 g


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