Question

The radioactive isotope 32P decays by first-order kinetics and has a half-life = 14.3 days. How many atoms of a 1.0000 ug sample of 32P would decay in 1.0000 second? (Assume 1 decay per atom.)

Answer #1

for first oder kinetics= half life= 0.693/K

K= rate constant =0.693/ half lfie

14.3 days= 14.3days*24hrs/day*60minutes/hr* 60 seconds/minute=1235520 second

K= rate constant= 0.693/1235520=5.6089*10-7/sec

mass of sample = 1ug= 1*10-3 gms

at t=1 sec

moles of 32P in1ug= 1*10-3/32=0.00003125

atoms in 0.0000 3125 = 0.00003125*6.023*10^{23}
atoms=1.882*10^{19} atoms

Atoms present at t= 1sec

Nt=
1.882*1019*exp(^{-5.6089*10-7*1)=}0.9999*1.882*10^{19}
atoms

Atoms that have undergones decay= initial atoms- atoms at time t=1

1.882*10^{19}-0.9999*1.882* 10^{19}
=10^{19}*(1-0.9999)=1*10^{-4}*10^{19} =
10^{15} atoms

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