Polonium 210 decays to Lead by alpha decay. At time t = 0 seconds, there are 150

Question

Polonium 210 decays to Lead by alpha decay. At time t = 0 seconds, there are 150 grams of pure Polonium 21084Po. The half-life of 21084Po is 138.4 days. The alpha decay reaction can be written as 21084Po ? 20682Pb + 42He a. Find the number of Polonium 210 nuclei present at initial time t = 0; assume one mole of Polonium has mass of 210 grams. b. Find the initial activity of the Polonium 210 sample in decays/second. c. Find the number of Polonium 210 nuclei at time t = 200 days. d. Find the total energy released in the alpha decay, expressed in MeV. The following are the masses: Polonium 21084Po : 209.982874 u Lead 20682Pb : 205.974465 u Helium 42He : 4.002603 u

Explanation / Answer

a. no of Polonium 210 nuclei present at initial time t = 0 is (150/210)*6.023*10^(23) = 4.3 *10^(23)

b. lambda = (half-life)^(-1) = 8.36*10^(-8) sec^(-1)

     so activity at t=0 is N0*lambda = 3.6 * 10^(16) decays/second

c. now we know N = N0 * e^(-lambda*t) = 4.3 *10^(23) * e^(-8.36*10^(-8)* 200*24*3600)

                                  = 1.014 *10^(23) nuclei

d. Energy released = total mass of reactants - total mass of products = (209.982874 - 205.974465 -

                                                                                                                         4.002603) * 931.46 MeV

                                   = 5.408 MeV

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