The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accide
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Question
The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level accepatable for humans.How long will it be until the region is habitable?
Please show work. The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level accepatable for humans.
How long will it be until the region is habitable?
Please show work.
Explanation / Answer
get the idea from this... According to Newton’s law of cooling the time rate of change of the temperature T(t) of a body immersed in a medium of constant temperature A is proportional to the difference A - T. That is dT dt = k(A - T), where k is positive constant. Example 1. A pitcher of butter milk initially at 250C is to be cooled by setting it on the front porch, where the temperature is 00C. Suppose that the temperature of the buttermilk has dropped to 150C after 20 min. When it will be at 50C? Solution. We note that A = 0, T(0) = 25, T(20) = 15. Denote by t0 the time moment when T(t0) = 5. Then by Newton’s law dT dt = -kT. The general solution to this equation is T(t) = Ce-kt. Thanks to the initial condition T(0) = 25 we have C = 25 and T(t) = 25e-kt. Next we find the constant k. Since T(20) = 15 = 25e-20k we have k = ln 53 20 . The formula for the temperature is T(t) = 25e- ln 5 3 20 t. Now we are able to find t0: T(t0) = 5 = 25e- ln 5 3 20 t0 . From this equation we obtain immediately t0 = 20 ln 5 ln 5 3 . Typeset by AMS-TEX 1 Population Growth. Suppose that P(t) is the number of individuals in a population having constant birth and death rates and . Then the dynamics of a population can be described by the ordinary differential equation dP dt = kP(t) where k = - . Example 2.In a certain culture of bacteria the number of bacteria increased sixfold in 10 h. How long did it take for the population to double? Solution. Let P(t) be a number of bacteria at moment t. Then dP dt = kP Denote by P0 the number of bacteria at moment t = 0. Then P(t) = P0ekt. Since P(10) = 6P0 we have 6P0 = P0e10k This formula implies immediately that k = ln 6 10 . Hence P(t) = P0e ln 6 10 t Let t0 be a moment of time such that P(t0) = 2P0. Then 2P0 = P0e ln 6 10 t0 and t0 = 10 ln 2 ln 6 . Radioactive decay. Consider the sample of a certain material that contains N(t) atoms of a certain radioactive isotope at time t. It has been observed that a constant fraction of those radioactive atoms will spontaneously decay during each unit of time. We describe this process by the ordinary differential equation dN dt = -kN. Here positive constant k depends on particular radioactive isotope. 2 Example 3.The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable? Solution. Denote by N(t) the mass of radioactive cobalt at moment t. Then dN dt = -kN If N0 is the mass of radioactive cobalt at moment t = 0 then N(t) = N0e-kt Since N(5.27) = N0 2 we have N(5.27) = N0 2 = N0e-5.27k. Then k = ln 2 5.27 . Hence N(t) = N0e- ln 2 5.27 t The region will be habitable when only 99 procent of cobalt decay. Denote by t0 the time moment when this happen. Then N(t0) = N0 100 = N0e- ln 2 5.27 t0 This formula implies t0 = 5.27 ln 100 ln 2 . Torricelli’s law Suppose a tank with water has a hole at its bottom. Denote by y(t) the level of a water in the tank at moment t. Then the function y(t) satisfies the ordinary differential equation A(y(t))dy dt = -k p y, Here A(y) is the horizontal cross sectional area of the tank and k is some positive constant. Example 4. At time t = 0 the bottom plug of a full conical water tank 16 ft high is removed. After one hour the water in the tank is 9 ft deep. When the tank will be empty? 3 Solution. Denote by y(t) the height of a water in a tank. Then A(y(t))dy dt = -k p y Note that A(y(t)) = r(t)2 = Ly2(t) Then y2 dy dt = µ p y, where µ = -k/(L). This equation is separable,so y 5 2 (t) = µt + C Using the initial condition y(0) = 16 we obtain y 5 2 (0) = 45 = C So y 52 (t) = µt + 45. Since y(1) = 9 we have y 5 2 (1) = 35 = µ + 45 We have µ = 35 - 45 Denote by T0 the time moment when y(t0) = 0 Then y 52 (t0) = 0 = (35 - 45)t0 + 45. Hence t0 = 45 45 - 35 .
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