Question

In March 2011, the Fukushima Daiichi nuclear power plant in Japan experienced a meltdown; consequently, radioactive materials were released into the environment. On the international Nuclear Event Scale (INES) running from 0 (indicating an abnormal situation with no safety consequences) to 7 (indicating an accident causing widespread contamination with serious health and environmental effects), the incident received a rating of 7. The only other recorder level 7 incident is the Chernobyl (Ukraine) meltdown in 1986.

18. One of the radioactive materials released during the Fukushima disaster was cesium-134; cesium-134 has a half-life of 2.1 years. How long will it take for this material to lose the following amount of its radioactivity?

a. 90%

b 99%

c. 99.9%

Answer 1

We know that radioactive decay is of the form :

dA/dt = -kA where A is the amount of radioactive material at time t and k is a constant , let A0 be the initial amount

Now dA/dt = -kA => dA/A = - kdt => ln(A/A0 ) = -k(t -0) => ln(A/A0) = -kt

half life is when A = A0/2 => ln(A0/2A0) = - kt_1/2 => k = ln2/t_1/2

using half life = 2.1 year , k = ln2 / 2.1 = 0.693/2.1 => k = 0.33 year^-1

(a) A = 90% lost of A = 10% remaining of A0 = 0.1A0

Now ln(0.1A0/A0) = -k*t => -ln(10) = -0.33* t => t = ln(10)/0.33 = 6.977 years

(b) A = 99% lost of A = 1% remaining of A0 = 0.01A0

Now ln(0.01A0/A0) = -k*t => -ln(100) = -0.33* t => t = 2ln(10)/0.33 = 2*6.977 years = 13.955 years

(c) A = 99.9% lost of A = 0.1% remaining of A0 = 0.001A0

Now ln(0.001A0/A0) = -k*t => -ln(1000) = -0.33* t => t = 3ln(10)/0.33 = 3*6.977 years = 20.932 years