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Find The age for a rock for which you determine that 65% as the original uranium – 238 remainswhile the other 35% has decayed into lead

Guest Jul 29, 2015

Best Answer 

 #2
avatar+18715 
+8

Find The age for a rock for which you determine that 65% as the original uranium – 238 remainswhile the other 35% has decayed into lead

 

I assume: uranium-lead decay route $${}^{238}U$$ to $${}^{206}{Pb}$$

 

(i)

$$\boxed{~N_t = N_0\cdot e^{-k\cdot t} ~~(1)~} \quad \small{\text{k, is called the decay constant.}}\\ \small{\text{$N_0$ is the initial number of nuclei (at time zero), and}} \\ \small{\text{$N_t$ is the number remaining after the time interval.}} \\ \small{\text{We can also write our equation:~}}
\boxed{~\ln{ \left( \dfrac{N_t}{N_0} } \right) = -k\cdot t ~~(2)~}$$

(ii)

$$\small{\text{Relationship between the decay constant, k, and half-life, $t_{1/2}$.}}\\
\small{\text{We set $N_{t_{1/2}} = \dfrac{N_0}{2} $ in (1)~so we have $ \dfrac{N_0}{2}=N_0\cdot e^{-k\cdot t_{1/2}} $ and get }}\\
\boxed{~ k = \dfrac{ \ln{(2)} }{ t_{1/2} } ~~(3)~}$$

(iii)

$$\small{\text{We get t, if we rearrange (2). We have~~}}
\boxed{~ t = -\dfrac{1}{k} \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(4)~}\\
\small{\text{We set k in (3) into (4). We have~~}}
\boxed{~ t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(5)~}\\$$

(iv)

$$\small{\text{$N_0 = {}^{238}U + {}^{206}Pb\cdot \dfrac{238}{206} $. And we have $35\%$ of ${}^{238}U$ is ${}^{206}Pb$ so }}\\
\small{\text{$N_0 = {}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}. $We obtain the latter quantity of ${}^{206}Pb$ by}}\\
\small{\text{multiplying the present mass of lead-206 by the ratio of the atomic mass of uranium}}\\
\small{\text{to that of lead, into which it has decayed. $N_t ={}^{238}U$. }}$$

$$\small{\text{ Finally
$\dfrac{N_t}{N_0} = \dfrac{ {}^{238}U } {{}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
$
}}\\
\boxed{ \dfrac{N_t}{N_0} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
}$$

(v)

$$\small{\text{If we regard the half-life $t_{1/2}$ for the decay of}}\\
\small{\text{uranium-238 $({}^{238}U)$ to lead-206 $({}^{206}Pb)$ is $\mathbf{4.468\cdot 10^9~ yr }$, }}\\
\small{\text{we can calculate how old the rock is.}}\\
\small{\text{We now have
$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}} \\
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ \ln{ (1) } -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\
}}\\
\small{\text{Because $\ln{(1)}= 0$
}} \\
\small{\text{$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ 0 -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\\\
\small{\text{and finally}}\\\\
\boxed{t = t_{1/2} \cdot \dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }}$$

(vi)

$$\small{\text{We calculate:}}\\
\small{\text{$
\begin{array}{rcl}
t &=&
t_{1/2} \cdot
\dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ \ln{ \left( 1.40436893204} \right) }
{ \ln(2) }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ 0.33958804319 }
{0.69314718056 }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot 0.48992198586 \\\\
\mathbf{t} &\mathbf{=}& \mathbf{ 2.18897143282\cdot 10^9~ yr }
\end{array}
$}}$$

 

The rock is 2 188 971 432.82 years old 

 

see SAMPLE EXERCISE 21.7 : http://wps.prenhall.com/wps/media/objects/3313/3392670/blb2104.html 

 

heureka  Jul 29, 2015
Sort: 

5+0 Answers

 #1
avatar+26329 
+5

The radioactive decay equation is N/N0 = e-ln(2)*t/τ, where N/N0 is the fraction at time t, and τ is the half-life. If we want to find t, we can take logs of both sides and rearrange to get:  t = -ln(N/N0)*τ/ln(2)

 

U238 has τ = 4.468*109 years, so here we have 

 

$${\mathtt{t}} = {\mathtt{\,-\,}}{\frac{{ln}{\left({\mathtt{0.65}}\right)}{\mathtt{\,\times\,}}{\mathtt{4.468}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{9}}}}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{t}} = {\mathtt{2\,776\,810\,067.302\: \!335\: \!197\: \!609\: \!348\: \!8}}$$

 

or t ≈ 2776810067 years.

.

Alan  Jul 29, 2015
 #2
avatar+18715 
+8
Best Answer

Find The age for a rock for which you determine that 65% as the original uranium – 238 remainswhile the other 35% has decayed into lead

 

I assume: uranium-lead decay route $${}^{238}U$$ to $${}^{206}{Pb}$$

 

(i)

$$\boxed{~N_t = N_0\cdot e^{-k\cdot t} ~~(1)~} \quad \small{\text{k, is called the decay constant.}}\\ \small{\text{$N_0$ is the initial number of nuclei (at time zero), and}} \\ \small{\text{$N_t$ is the number remaining after the time interval.}} \\ \small{\text{We can also write our equation:~}}
\boxed{~\ln{ \left( \dfrac{N_t}{N_0} } \right) = -k\cdot t ~~(2)~}$$

(ii)

$$\small{\text{Relationship between the decay constant, k, and half-life, $t_{1/2}$.}}\\
\small{\text{We set $N_{t_{1/2}} = \dfrac{N_0}{2} $ in (1)~so we have $ \dfrac{N_0}{2}=N_0\cdot e^{-k\cdot t_{1/2}} $ and get }}\\
\boxed{~ k = \dfrac{ \ln{(2)} }{ t_{1/2} } ~~(3)~}$$

(iii)

$$\small{\text{We get t, if we rearrange (2). We have~~}}
\boxed{~ t = -\dfrac{1}{k} \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(4)~}\\
\small{\text{We set k in (3) into (4). We have~~}}
\boxed{~ t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(5)~}\\$$

(iv)

$$\small{\text{$N_0 = {}^{238}U + {}^{206}Pb\cdot \dfrac{238}{206} $. And we have $35\%$ of ${}^{238}U$ is ${}^{206}Pb$ so }}\\
\small{\text{$N_0 = {}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}. $We obtain the latter quantity of ${}^{206}Pb$ by}}\\
\small{\text{multiplying the present mass of lead-206 by the ratio of the atomic mass of uranium}}\\
\small{\text{to that of lead, into which it has decayed. $N_t ={}^{238}U$. }}$$

$$\small{\text{ Finally
$\dfrac{N_t}{N_0} = \dfrac{ {}^{238}U } {{}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
$
}}\\
\boxed{ \dfrac{N_t}{N_0} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
}$$

(v)

$$\small{\text{If we regard the half-life $t_{1/2}$ for the decay of}}\\
\small{\text{uranium-238 $({}^{238}U)$ to lead-206 $({}^{206}Pb)$ is $\mathbf{4.468\cdot 10^9~ yr }$, }}\\
\small{\text{we can calculate how old the rock is.}}\\
\small{\text{We now have
$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}} \\
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ \ln{ (1) } -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\
}}\\
\small{\text{Because $\ln{(1)}= 0$
}} \\
\small{\text{$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ 0 -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\\\
\small{\text{and finally}}\\\\
\boxed{t = t_{1/2} \cdot \dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }}$$

(vi)

$$\small{\text{We calculate:}}\\
\small{\text{$
\begin{array}{rcl}
t &=&
t_{1/2} \cdot
\dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ \ln{ \left( 1.40436893204} \right) }
{ \ln(2) }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ 0.33958804319 }
{0.69314718056 }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot 0.48992198586 \\\\
\mathbf{t} &\mathbf{=}& \mathbf{ 2.18897143282\cdot 10^9~ yr }
\end{array}
$}}$$

 

The rock is 2 188 971 432.82 years old 

 

see SAMPLE EXERCISE 21.7 : http://wps.prenhall.com/wps/media/objects/3313/3392670/blb2104.html 

 

heureka  Jul 29, 2015
 #3
avatar+91045 
0

Two great answers, thanks Alan and Heureka  

Melody  Jul 29, 2015
 #4
avatar+26329 
+5

I think heureka has been misled (in his section (iii)) by the example at : http://wps.prenhall.com/wps/media/objects/3313/3392670/blb2104.html 

 

We are told directly here that Nt/N0 = 0.65, so that's all we need.

 

If the mass of Uranium now is Nt and the original mass was N0 then heureka's formula for original mass should just be:

N0 = Nt + 0.35*N0   which results in N0 = Nt/(1-0.35), so Nt/N0 = (1-0.35) = 0.65.

We don't need the relative atomic masses of U238 and lead206 (we would do if we had the actual masses of U238 and Pb206 now).

 

heureka's expression for Nt/N0 gives a ratio of just over 71%

.

Alan  Jul 29, 2015
 #5
avatar+91045 
0

You are probably right Alan but you have to admit, Heureka's answer LOOKED very impressive.   LOL

 

Heureka's answers are always impressive   

Melody  Jul 29, 2015

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