Nuclear Chemistry and Half-life calculations?

If Rb-87 decays to form Sr-87 and the half-life of the reaction is 4.8E^9, what is the nuclear equation of the reaction? Also if .100% of a rock is Rb-87, and .00250% is Sr-87, assuming the rock did not contain any Sr-87 beforehand. How old is the rock? Would the answer be around 2.57E^11 years old?

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3 Answers
Doc048
May 22, 2018

Age of rock = #1.71xx10^8 color(white)(x)yearscolor(white)(x)old#

Explanation:

#""_37^87Rb# => #""_38^87Sr# + #beta^-#

Given:
Assuming rock did not contain Sr-87 at start (from problem text), and all Sr-87 comes from decay of Rb-87, then...

#A_o=%(Rb87) + %(Sr87)# = #0.100% + 0.00250% = 0.10250%#

#A_"final" = %Rb87color(white)(x)"at"color(white)(x) "time" (t) = 0.100%"#

#t_(1/2)#=#4.8xx10^9yrs# => #k=0.693/t_(1/2)#=#0.693/(4.8xx10^9)yrs^-1#=#1.444xx10^-10yrs^-1#

Apply to 1st order decay equation #A_"final" = A_oe^-"kt"#
and solve for time (t) => #t = (ln(A_f/A_o))/(-k)# = #ln((0.100)/(0.10250))/(-1.444xx10^-10)yrs#= #1.71xx10^8yrs#

A08
May 23, 2018

Nuclear equation for Rubidium-87 decay to Strontium-87 by beta decay is

#""_87^37"Rb" -> ""_87^38"Sr" +beta_0^(-1)#

Given initial sample

Has #"Rb"# only
After time #t#, ratios are#=0.100%\ "Rb"# and # = 0.00250%\ "Sr"#
#t_(1/2)=4.8xx10^9yrs#
#=> lambda=ln2/t_(1/2)=ln2/(4.8xx10^9)=1.444xx10^-10\ yrs^-1#

The amount of #87"Sr"# found in a sample at any time is determined by

  1. the decay constant of #87"Rb"#,
  2. the initial amount of #87"Sr"# in the sample,
  3. the time elapsed since the initial time and
  4. the ratio of #"Rb"# to #"Sr"# in the system.
    This can be seen in the equation below.

#87"Sr"("sample")= 87"Sr"("initial")+87"Rb"("sample")(e^(lambdat)-1)#
where #lambda# is the decay constant and #t# is the age of the sample.

Inserting given values we get

#0.00250= 0.100(e^(1.444xx10^-10t)-1)#
#=> (e^(1.444xx10^-10t)-1)=0.00250/0.100#
#=> e^(1.444xx10^-10t)=1.025#

Taking natural logarithm of both sides and solving for #t# we get

#t=1.71xx10^8\ yrs#

Michael
May 23, 2018

#sf(1.71xx10^8)# years

Explanation:

Rubidium undergoes #sf(beta)# decay:

#sf(""_37^87Rbrarr)##sf(""_38^(87)Sr)#+#sf(beta)#

At the start of the decay process the sample only contains Rb - 87. As it decays exponentially the amount of Sr - 87 grows correspondingly.

The decay is a 1st order process such that:

#sf(Rb_t=Rb_0e^(-lambdat))#

#sf(Rb_0)# is the number of undecayed atoms of Rb - 87 initially.

#sf(Rb_t)# is the number of undecayed atoms of Rb - 87 after time t.

#sf(lambda)# is the decay constant.

Since the decay of 1 atom of Rb - 87 produces 1 atom of Sr - 87 we can say that:

#sf(Rb_0=Rb_t+Sr_t)#

Where #sf(Sr_t)# is the number of undecayed atoms of Sr - 87 formed after time t.

The decay equation can therefore be written:

#sf(Rb_t=(Rb_t+Sr_t)e^(-lambdat))#

#:.##sf((Rb_t)/((Rb_t+Sr_t))=e^(-lambdat))#

#:.##sf(((Rb_t+Sr_t))/(Rb_t)=e^(lambdat))#

#:.##sf(1+(Sr_t)/(Rb_t)=e^(lambdat))#

#:.##sf((Sr_t)/(Rb_t)=e^(lambda)-1)#

The half - life of Rb - 87 is #sf(4.8xx10^(9))# years.

We can get the value of the decay constant from the expression:

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

#sf(lambda=0.693/(4.8xx10^(9))=1.44xx10^(-10)color(white)(x)yr^(-1))#

We can get the number of moles of each isotope by dividing the % mass by the mass of 1 mole:

#sf(n_(Rb)=0.100/87)#

#sf(n_(Sr)=0.00250/87)#

There is no need to convert these into a number of atoms by multiplying these by the Avogadro Constant since we are interested in the Sr/Rb ratio so this cancels anyway.

#:.##sf(0.00250/87/0.1/87=e^(lambdat)-1)#

#:.##sf(0.0250=e^(lambdat)-1)#

#sf(e^(lambdat)=1.0250)#

Taking natural logs of both sides #sf(rArr)#

#sf(lambdat=ln(1.0250)#

#sf(t=0.02469/(1.44xx10^(-10))=1.71xx10^(8))# years

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