radioactive decay usually follows a first order rate law. The half life for 239-

Question

radioactive decay usually follows a first order rate law. The half life for 239-Pu is 24,400 years and the half-life for the decay of 241-Pu is 13.0 years.

a) calculate the rate constants for the decay of 239-Pu and 241-Pu. which decays more rapidly?

b) if one starts with 5 g of pure 241-Pu, how many disintergrations per second occur initially? what mass of 241-Pu is left after 1 year? 10 years? 100 years?

c) 239-Pu can be used to synthesize 241-Am by bombarding the 239-Pu nucleus with two neutrons; 241- Pu is an intermediate. Show the complete equations for synthesis of 241-Am

Detailed walkthrough of all please!

Thanks!

Explanation / Answer

a)

For 239-Pu,

Q = Q_0*e^(-rt)

half-life is 24,400 years

so, Q_0/2 =Q_0*e^(-r*24400 )

rate constants for the decay of 239-Pu = r= 0.00002840767 year ^-1 = 0.0000000000009008 s^-1

For 241-Pu ,

Q = Q_0*e^(-rt)

half-life is 13 years

so, Q_0/2 =Q_0*e^(-r*13 )

rate constants for the decay of 241-Pu = r= 0.0533190 year ^-1 = 0.0000000016907344 s^-1

b)

5 g of pure 241-Pu initially ,

Q_0 = 5 gm

Number of moles of 241-Pu = 5/241 = 0.020746887 moles

Number of molecules = 6.023*10^23 * 0.020746887 = 0.124958500401 *10^23

how many disintergrations per second occur initially = N_0*r =0.124958500401 *10^23* 0.0000000016907344 = 21.12716352 *10^12 Bq (Bq=1desintigration/s) is the answer

mass of 241-Pu is left after 1 year = 5* e^(-0.0533190*1) = 4.740387637536175 gm

mass of 241-Pu is left after 1 year = 5* e^(-0.0533190*10) =2.9336515574730485 gm

mass of 241-Pu is left after 1 year = 5* e^(-0.0533190*100) = 0.0241743750861856 gm

c)

bombarding the 239-Pu nucleus with two neutrons

239-Pu + 2 N --> 241- Pu

where N represents neutrons

241- Pu ---> 241 - Am + B

where B represents beta particle

so, the whole synthesis is

239-Pu + 2 N --> 241- Pu

241- Pu ---> 241 - Am + B

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