Name: Liquid Drop Model of Nucleus
Uploaded: 2019-07-19
Duration: 2 min 42 s
Description: Did you know a liquid drop shares many similarities with a nucleus? It will be interesting to learn the similarities let's learn the liquid drop model of the...

Question

U^{238}

and

U^{235}

occur in nature in an atomic ratio

140 : 1

. Assuming that at the time of earth's formation the two isotopes were present in equal amounts. Calculate the age of the Earth. (Half life of

U^{238} = 4.5 \times 10^{9}

yrs & that of

U^{235} = 7.13 \times 10^{8}

yrs )

Accepted Answer

We use the formula

$C_{t} = C_{0} e^{- kt} . . . . . (1)$

For first order reaction, $k = \frac{In 2}{T} . . . . . (2)$

Where, $C_{t}$ is concentration of reactant at time tt, $C_{0}$ is initial concentration of reactant, $t$ is time, $k$ is rate constant and $T$ is half-life of the reactant.

We are given the question that initial concentrations of $U^{238}$ and $U^{235}$ are equal.

Let the initial concentrations of $U^{238}$ and $U^{235}$ be $C_{0}$ and final concentrations of $U^{238}$ and $U^{235}$ at time $t$ be $C_{1}$ and $C_{2}$ respectively.

We know that $C_{t} = C_{0} e^{- kt}$ .

Hence,

$\Rightarrow C_{1} = C_{0} e^{- k_{1} t} . . . . . (3)$

$\Rightarrow C_{2} = C_{0} e^{- k_{2} t} . . . . . (4)$

Where $k_{1}$ and $k_{2}$ are the rate constants for the decomposition of $U^{238}$ and $U^{235}$ respectively.

We know that, $\frac{C_{1}}{C_{2}} = \frac{140}{1} . . . . . (5)$

Dividing equation $(3)$ by equation $(4)$ , we get

$\frac{C_{1}}{C_{2}} = \frac{C_{0} e^{- k_{1} t}}{C_{0} e^{- k_{2} t}} = \frac{e^{- k_{1} t}}{e^{- k_{2} t}}$

$\frac{C_{1}}{C_{2}} = e^{- k_{1} t - k_{2} t}$

Taking natural log on both sides of the equation, we get

$(k_{2} - k_{1}) t = \ln (\frac{C_{1}}{C_{2}})$

Rearranging to get $t$ ,

$t = \frac{\ln (\frac{C_{1}}{C_{2}})}{(k_{2} - k_{1})} . . . . . (6)$

From equation $(2)$ , we know that,

$k_{1} = \frac{\ln 2}{T_{1}}$ and $k_{2} = \frac{\ln 2}{T_{2}}$ where $T_{1} = 4.5 \times 10^{9} years$ and $T_{2} = 7.13 \times 10^{8} years$ are half-lives of $U^{238}$ and $U^{235}$ respectively.

Substituting these values in equation $(6)$ , We get,

$\Rightarrow t = \frac{\log (140)}{\log 2} [\frac{T_{1} \times T_{2}}{T_{1} - T_{2}}] \Rightarrow t = \frac{In (\frac{C_{1}}{C_{2}})}{In 2 (\frac{1}{T_{1}} - \frac{1}{T_{2}})} = \frac{\log (140)}{\log 2 (\frac{T_{1} - T_{2}}{T_{1} . T_{2}})}$

$\Rightarrow t = \frac{\log (140)}{\log 2} [\frac{T_{1} \times T_{2}}{T_{1} - T_{2}}]$

$\Rightarrow t = \frac{2.1461}{0.3010} [\frac{4.5 \times 10^{9} \times 7.13 \times 10^{8}}{4.5 \times 10^{9} - 7.13 \times 10^{8}}]$

$∴ t = 6.04 \times 10^{9} years$

Hence, it is the required answer.

Embibe Experts Solutions for Chapter: Nuclear Physics, Exercise 4: Exercise-4

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Nuclear Physics, Exercise 4: Exercise-4

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