+0

# 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

0
319
5

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
+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
+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
+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
+91045
0

Two great answers, thanks Alan and Heureka

Melody  Jul 29, 2015
#4
+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
+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

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details