Question

Only problem 8 please.

8. Consider the three component decay in Problem 7. The half-life of the orig- inal nucleus is one day. The half-life of the second nuclide is two days, while the third nuclide is stable. The initial number of particles for nuclide A is 10,000, while nuclides B and C have no initial particles. Plot the number of particles from 0 to 10 days using ODE45 in MATLAB.

Problem 7 Consider the case where we have the three component decay chain A C(Stable) Derive the formula for the number of particles of B over time, proving the answer is -e-ABt

Answer 1

i have derived the expression for number of B particles as a function of time and then plotted the function ffor time = 0 to 10 days in matlab whose screenshots are attached

(t_v_2)_A = 1 day tv_2 = ln 2/lambda = 0.693/lambda (tv_2)_B = 2 days (N_0)_A = 1000 (N_0)_B = (N_0)_C = 0 No of B form per sec = NO of A decay per sec d N_A/dt = -lambda_a N_A implies N_A (t) = N_0 (A) e^-lambda_A t d N_B/dt = lambda_A N_A -lambda_B N_B implies N_B (t) = f(t) e^-lambda_B t dN_B/dt = f(t)(-lambda_B e^-lambda_B Bt) + df(t)/dt) e^-lambda_B t = lambda_A N_A - lambda_B N_B implies df(t)/dt = lambda_A (N_0)_A e^-lambda_A t e^lambda_B t implies integral f(t) = integral lambda_A (N_0)_A e^(-lambda_A t) e^lambda_B t implies f(t) = lambda_A (N_0)_A e^-lambda_A - lambda_B) t/lambda_B - lambda_A at t = 0, (N_0)_B = 0 implies f(t) = 0 implies c = -lambda_A (N_0)_A/lambda_B - lambda_A f(t) = lambda_A(N_0)_A/(lambda_B - lambda_A) [e^-lambda_A t - e^-lambda_B t) N_B (t) = f(t) e^-lambda_B t = lambda (N_0)t/lambda_B - lambda_A [e^-lambda_A t - e^-lambda_B t]
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