Please show your solution on paper using the formulas in the picture. Thank you so much!Question: The Fukushima nuclear disaster in Japan is the most recent catastrophic nuclear disaster in memory which was brought about by the massive tsunami that was triggered by a magnitude 9.0 earthquake in Tohoku region in Japan last 2011. One of the radioactive substances released during the incident is cesium-137 which has a half-life of 30 years. Around 15,000 terabecquerels (becquerel is a standard unit of radioactivity named after Henri Becquerel) of Cs-137 was released during the disaster. How many becquerels of this radioactive substance remains to this day? The rate of radioactive decay follows first-order kinetics where the rate of decay isproportional to the number of radioactive nuclei (N) in the sample as expressed in Equation 6.1where k is called the nuclear decay constant.Rate = kN (Equation 6.1)Equation 6.1 can be transformed into Equation 6.2 where No is the initial number of nuclei at initialtime, time =0, and N, is the number of nuclei after time a certain time interval, t.NtIn =No-kt (Equation 6.2)A more useful way of determining the rate of radioactive decay is by determining the half-life of a radioisotope. Half-life (t₁/2) is the time required for half of any given quantity of radioactivesubstance to decay. Each radioisotope has a characteristic half-life. For example, cobalt-60which is used for cancer radiation therapy has a half-life of 5.3 years. So for a 1.00 g sample ofcobalt-60 it will take 5.3 years before its amount is reduced to 0.500 g and 10.6 yrs to 0.250g andso on and so forth. Equation 6.3 gives the general equation for the half-life of any radioactivesubstance while Equation 6.4 is the general formula in calculating the amount of remainingsubstance after n half-lives.t1/2 = (Equation 6.3)0.693kNt= (1/2) No (Equation 6.4)

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Please show your solution on paper using the formulas in the picture. Thank you so much!

Question: The Fukushima nuclear disaster in Japan is the most recent catastrophic nuclear disaster in memory which was brought about by the massive tsunami that was triggered by a magnitude 9.0 earthquake in Tohoku region in Japan last 2011. One of the radioactive substances released during the incident is cesium-137 which has a half-life of 30 years. Around 15,000 terabecquerels (becquerel is a standard unit of radioactivity named after Henri Becquerel) of Cs-137 was released during the disaster. How many becquerels of this radioactive substance remains to this day?

The rate of radioactive decay follows first-order kinetics where the rate of decay is
proportional to the number of radioactive nuclei (N) in the sample as expressed in Equation 6.1
where k is called the nuclear decay constant.
Rate = kN (Equation 6.1)
Equation 6.1 can be transformed into Equation 6.2 where No is the initial number of nuclei at initial
time, time =0, and N, is the number of nuclei after time a certain time interval, t.
Nt
In =
No
-kt (Equation 6.2)
A more useful way of determining the rate of radioactive decay is by determining the half-
life of a radioisotope. Half-life (t₁/2) is the time required for half of any given quantity of radioactive
substance to decay. Each radioisotope has a characteristic half-life. For example, cobalt-60
which is used for cancer radiation therapy has a half-life of 5.3 years. So for a 1.00 g sample of
cobalt-60 it will take 5.3 years before its amount is reduced to 0.500 g and 10.6 yrs to 0.250g and
so on and so forth. Equation 6.3 gives the general equation for the half-life of any radioactive
substance while Equation 6.4 is the general formula in calculating the amount of remaining
substance after n half-lives.
t1/2 = (Equation 6.3)
0.693
k
Nt= (1/2) No (Equation 6.4)

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