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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

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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

#2
+19054
+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
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#1
+26544
+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
+19054
+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
#3
+91900
0

Two great answers, thanks Alan and Heureka

Melody  Jul 29, 2015
#4
+26544
+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
+91900
0

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