Appm 2360 Radioactive Decay Project Assessment Answers

Questions:

Answer:

The rate of change of particles in radioactive decay is represented by,

Rate of change = Rate in – Rate out

In the initial state as no new particles are created hence the equation becomes

Rate of change = -Rate out

Now, in mathematical notation 

Here, = the number of atoms present in a given time t of the decay element A.

 = rate constant.

Question 2.2:

• Now, the given differential equation is 

With initial condition =

Now, rearranging the differential equation and then integrating both sides gives,

 = = ln(

Now, putting the initial condition gives,

ln(

hence, the solution of the differential equation is

ln(

• a) Now, the result of the question 1 to solve for is given by, 

1. b) Now, given that = (½)

Hence, at half time the expression of will be,

 = ln(1/2)/

1. c) Now, given that the half life time is 

So, the rate constant at half lifetime is

ln(1/2)/(5ln(2)) = -0.2

Hence, from b) the unit of the proportionality constant is time^(-1) or sec^(-1).

Hence, = -0.2

1. Now, the proportional rate = -0.2 is a constant and hence the differential equation becomes 

With initial condition = 15000 atoms.

Now, the exact solution of the equation is,

 = exp(-0.2*t) => = 15000*(exp(-0.2*t))

Now, the exact solution along with the Euler method solution for step sizes h=1,0.1,0.01 is plotted in a same graph by the following MATLAB code.

MATLAB code:

function euler

h1=1;h2=0.1;h3=0.01;

N=40;

NA1(1)=15000;NA2(1)=15000;NA3(1)=15000; % defining the initial number of atoms

% Euler for h=1

for n=1:N

NA1(n+1)=NA1(n)+h1*(-0.2*NA1(n)); % defining the differntial equ dNA/dt = -0.2NA

t1(n+1)=n*h1;

end

% Euler for h=0.1

for n=1:N

NA2(n+1)=NA2(n)+h2*(-0.2*NA2(n)); % defining the differntial equ dNA/dt = -0.2NA

t2(n+1)=n*h2;

end

% Euler for h=0.01

for n=1:N

NA3(n+1)=NA3(n)+h3*(-0.2*NA3(n)); % defining the differntial equ dNA/dt = -0.2NA

t3(n+1)=n*h3;

end

exact = 15000*(exp(-0.2.*t1)); % exact solution 

plot(t1,NA1,'g-',t2,NA2,'m-',t3,NA3,'r-',t1,exact,'b-')

legend('Euler for h=1','Euler for h=0.1','Euler for h=0.01','exact solution')

title('Solution by Euler method of different step size and the exact solution')

xlim([0 0.4])

xlabel('time in sec')

ylabel('Atom population')

grid on

end

Plot: 

1. b) Absolute error plots for different step is obtained by adding the following code with the MATLAB code in part a.

MATLAB code:

err1 = abs(exact-t1);

err2= abs(exact-t2);

err3 = abs(exact-t3);

plot(t1,err1,'k-',t2,err2,'r-',t3,err3,'m-')

legend('error for h=1','error for h=0.1','error for h=0.01')

title('absolute errors for different step size')

grid on

xlabel('time in sec')

ylabel('absolute error')

Plot:

1. The numerical solution of the problem gives an approximate solution at a given time and as the step size becomes smaller the numerical solution reaches the exact solution but the time to reach the solution increases. Hence, the reasonable three step sizes that will gives acceptable efficiency (time required to reach the solution) as well as accuracy is h=1, 0.5 and 0.1.

• Now, ideally as the solution of the differential equation is exponential with negative power, hence the time when the there will be no atoms left in the sample of A is infinity, but from the plots in MATLAB, it can be seen that the time when the solution reaches to zero is approximately at time t = 40 secs. 

1. The rate of change of equation (3) is given by

N’(t) = Rate in – Rate out

Hence, 

Now, = A0*(exp(-*t)).

So, the differential equation becomes

(A)

1. Now, the solution of (A) can be found by multiplying it with the integrating factor on both sides.

Hence, 

Or, =

Or, = dt

Now, integrating both sides give,

 = Now, initial number of atoms are considered as 0 or = 0 or initially no atoms of A has broken into B.

Hence, 0 = + C => C =

Hence, =

Hence,

1. Now, the differential equation for in the similar way will be, 

Now, putting the solution of in the equation we have 

Or,

Now, multiplying both sides by the integrating factor gives 

Or, =

Now, integrating both sides gives,

 + C1

Now, considering initially there was no atoms of C or the decaying process is not started .

So, 0 = 1

Or, 0 = +C1

So, C1 =

Hence, =

1. Now, the half time of B is given as

 seconds

o, putting in

Now, the considering = k* = 0.2*2 = 0.4 sec^(-1) (k=2 as the initial concentration reduces the decay rate is increased)

Now, putting the values 

Similarly, = 0.6 sec^(-1).

So, =

= 

MATLAB code:

t = 0:1:40;

A0= 15000;

%%% rate A = 0.2, rate B = 0.4 and rate C=0.6

N2 = ((A0*0.2)/(0.4-0.2))*(exp(-0.2*t)-exp(-0.4*t));

N3 = (0.4*0.2*15000*exp(-0.6*t)/0.2).*(exp(0.4*t)/0.4 - exp(0.2*t)/0.2) + 0.4*0.2*15000/(0.4*0.2);

plot(t,N2,'r-',t,N3,'b-')

legend('number of atoms of B','Number of atoms of C')

title('The number of atoms of B and C')

xlabel('time in sec')

ylabel('number of atoms')

Plot: 

Now, in the above plot a local maximum is observed at about t = 5 secs of the atoms of B due to the rate in from A the number of atoms increase in B initially and then after some time the number of atoms in B starts to decrease.

5. Now, for the modified condition when element A is produced at the rate of P = 5000 atoms/sec the solution equation of ,and becomes like this

A0*(exp(-*t)) + 5000*t

 + 5000*t

 =And differential equations are

6. Now, using the Euler’s method with step size h=0.01 the above three differential equation is solved as given below.

MATLAB code:

function q6

h1=0.01;

N=100;

NA(1)=15000; % defining the initial number of atoms

% Euler for NA,NB and NC

for n=1:N

t1(n+1)=n*h1;

NA(n+1)=NA(n)+h1*(-0.2*NA(n)+5000*t1(n));

NB(n+1)=NA(n)+h1*(-0.4)*(-0.2*NA(n)+5000*t1(n));

NC(n+1)=NA(n)+h1*(-0.6)*(-0.4)*(-0.2*NA(n)+5000*t1(n));

end

7. Now, the plot of and together is obtained by the following MATLAB code in addition with the previous code.

plot(t1,NA,t1,NB)

legend('number of atoms of A','number of atoms of B')

Plot: 

The asymptotic behaviour of atoms of B is expected because of that the production rate of 5000 atoms/sec to A contributes in chain reaction of number of atoms in B and hence, a peak is observed in at first and then it slowly reduces asymptotically.

8. Now, the plot of Nc(t) is obtained by the following MATLAB code.

plot(t1,NC)

legend('number of atoms of C')

title('Nc(t)')

xlabel('time in sec')

ylabel('number of atoms')

Plot: 

Now, as t grows large it is seen that the number of atoms of C is slowly increases but reaches at a maximum of 15000 atoms as the number of atoms can not be larger than the initial number of atoms.

The additional conditions that may affect the radioactivity are the ambient temperature and the weather conditions atmospheric pressure. The decay of the radioactive atoms and the emissions of alpha, beta and gamma particles are accelerated in high temperature and pressure conditions.