Question

5) It is known that radioactive decay of a substance can be modeled as N = Noe"t/t, where No is the amount of substance N at time-0, and r is the mean lifetime of a radioactive particle before decay In le')= a) Linearize this equation. (4 points) N Ne ( b) Given the data in the table, perform the linear regression to the equation you wrote in part a). Show your work to receive full credit. (16 points) 2 h. -7 ao q.5 t (years) 0.5 1.5 2.0 2.5 1,0 Y 994 990 985 979 977 N Y 2.2 471 nExy- Ex Ey a1 nExf-Mx) ao 9-a, 7.5 t5 415 471 X Y 477 y7 615 4.S s(7369)-(7-90t9 116 1477.s 2.5 ai (13.1 1453 47f 2 6-25 2443- 77 y12 13-7 5 75-7 (.2) 7346 7-5 1.5 No c) Solve for T and No (4 points) 2558 yrer N N: 918 S 111 N:N N-NCO e) Use the model to predict the amount of material remaining after 20 years (2 points) X 2.5 a77y 2-5 N-977 a77 2.5 20 2 5977 417 2-5K11540 7 7

Answer 1

5

a) The given model is . To linearize this equation, take the logarithm on both sides

N(t) = Noet/7

using properties of logarithm.

n(Noet)In(No) +ln(e-t/) In N(t)

Simplifying further,

In N(t) In(No)

This is a linear equation for ln(N) in terms of t with slope $-\frac{1}{\tau}$ and intercept .

In(No 0

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b) Now perform a linear regression for the above equation. x in the formula for regression is time in years and y is ln(N), which we find from the given values of N.

x	0.5	1.0	1.5	2.0	2.5
y	ln 994	ln 990	ln 985	ln 979	ln 977

Then

$\sum x_i = 7.5 \\ \sum y_i = 34.46310 \\ (\sum x_i)^2 = 56.25 \\ \sum x_i^2 = 13.75 \\ \sum x_i y_i = 51.67181592 \\ \bar{x} = 1.5 \\ \bar{y} = 6.89262$

Using the above data and the given formulas, the slope is
-9.133632 x 10- a1
, and the intercept is
Lo6.906320448
.

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c) From a), we know that the intercept
a6.906320448 In(No)
. Taking exponential on both sides, we find
No 998.566
.

Similarly the slope $a_1 = -1/\tau = -9.133632 \times 10^{-3}$ . Taking inverse on both sides .

T=109.4854 7

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d) Substitute for $\tau, N_0$ in the model to find $N(t) = N_0 e^{-9.133632 \times 10^{-3}t }$ , where t is time in years.

The amount of material remaining after 20 years is N(20).

This is $N(20) = N_0 e^{-9.133632 \times 10^{-3} \times 20 } = 831.846$ .

After 20 years, the amount of material remaining is expected to be 831.846