Question

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?

Answer 1

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`

Answer 2

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`

Answer 3

`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