Question

Please show original and detailed work.

Consider the two step decay 23Ra-Rn-4Po. Both steps are alpha decays, with half-lives 1600 years forRa and 3.8235 days forRn (a) Determine the decay constant of each step in the chain (b) If you have a sample of pure radium-226 (238 Ra) containing 3.00x 1020 nuclei at time how many radon-2223Rn) nuclei are present at time t? (c) What is the maximum number of radon-222 Rn ) nuclei present at any time? (d) 5 pts Extra: Show graphically the number of 238Ra, 2Rn, and 214 Po nuclei present 86 function of time. You will need to think about appropriate scales for each graph to cl represent the functions. Comment on the shapes of the graphs

Answer 1

"Given that Consider two step decay R_a rightarrow R_n rightarrow P_o dn/dt = -lambda N(1): where 'N' is the number of nuclei that have not decayed yet, No is the initial number of nuclei and 'lambda' is the decay constant Besides, the half is related to the decay constant through the following relation T_1/2 = Ln(2)/lambda(2) (a) The decay constant of each step in the chain can be determined using equation (2) R_a: 1600 years = Ln(2)/lambda (R_a) ... lambda(Ra) = Ln(2)/(1600 years).... Lambda(Ra) = lambda_1 = 4.332 times 10^-4/years R_n: 3,8235 days = Ln(2)/lambda(R_n): : : lambda(R_n) = Ln(2)(3.8235 days): : : lambda(R_n) = lambda 2 = 0.1831/days (b) for this case we have a situation where the number of nuclei of R0 have gains (when the R_a days) and losses (when the R_n decays), therefore we have the following. dN^2/dt = lambda