Answer to Question #265374 in Differential Equations for mary

Question

1. Initially, there are 30 grams of Astatine, a highly radioactive element made in nuclear reactors. The half life of the element is 8 hours. Find the amount of substance remains for the first 5 hours. How long will it take for the 99.9% of its mass to decay? Ans. 19.45g; 79.73 hours
2.  The population of a community will be treble by the year 2009. if the population in the year 1998 initially has 2025, find the year when the population will be doubled. What is the population in the year 2012?  Ans. 2005; 8198

Answer 1

1.

$$R=R_0e^{kt}$$

A half life is 8 years

$$\dfrac{1}{2}R_0=R_0e^{k(8)}$$

$$k(8)=-\ln 2$$

$$k=-\ln 2/8$$

$$R=R_0(2)^{-t/8}$$

Initially $R_0=30 mg$.

(i)

$$R(5)=30(2)^{-5/8}$$

$$R(5)=19.45mg$$

(ii)

$$R=R_0(2)^{-t/8}=0.001R_0$$

$$\ln((2)^{-t/8})=\ln(0.001)$$

$$-t/8=\ln (0.001)/\ln(2)$$

$$t=8(3\ln(10)/\ln(2))$$

$$t=79.73\ years$$

2.

$$P=P_0e^{kt}$$

The population of a community will be treble for 11 years

$$3P_0=P_0e^{k(11)}$$

$$k(11)=\ln 3$$

$$k=\ln 3/11$$

$$P=P_0(3)^{t/11}$$

i)

$$P_1=2P_0$$

$$2P_0=P_0(3)^{t_1/11}$$

$$(3)^{t_1/11}=2$$

$$t_1=11\cdot\dfrac{\ln2}{\ln3}$$

$$t_1=6.94years$$

$$1998+6.94\approx2005$$

$2005$ year

ii)

Initially $P_0=2025$.

$$2012-1998=14$$

$$P_2=2025(3)^{14/11}$$

$$P_2=8198$$

The population is 8198 in the year 2012.