If the half life of uranium-235 is #7.04 * 10^8# y and 12.5 g of uranium-235 remain after #2.82 * 10^9# y, how much of the radioactive isotope was in the original sample?

1 Answer
Jan 20, 2016

Answer:

#200"g"#

Explanation:

1. Quick Method.

Examiners and text - books often give nice numbers where the answer drops out easily.

In this case you count the number of 1/2 lives #n_(1/2)# that have elapsed:

#n_(1/2)=(2.82xx10^9)/(7.04xx10^8)=4#

Now multiply back from 12.5 by doubling it successively 4 times:

Initial mass = #12.5 xx2xx2xx2xx2=200"g"#

1. General Method.

If the numbers are not so nice you need to use the general equation for radioactive decay:

#N_t=N_0e^(-lambdat)" "color(red)((1))#

#lambda# is the decay constant

#N_0# = the initial number of undecayed atoms

#N_t# = the number of undecayed atoms at time #t#

We can get #lambda# from the 1/2 life using this expression:

#lambda=0.693/(t_(1/2))#

#:.lambda=(0.693)/(7.04xx10^(8))=9.84xx10^(-10)" "a^(-1)#

Taking natural logs of both side of #color(red)((1))rArr#

#lnN_t=lnN_0-lambdat#

We can use mass in grams instead of atoms so:

#lnM_t=lnM_0-lambdat#

#:.lnM_0=lnM_t+lambdat#

#:.lnM_0=ln(12.5)+(9.84xx10^(-10)xx2.82xx10^9)#

#lnM_0=2.526+2.775=5.301#

#:.M_0=200.5"g"#