Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  A rock samples has ratios of atoms to <img alt="{ }^{206} \mathrm{~Pb}" src="/latex-image/?%7B%

 A rock samples has ratios of { }^{238} \mathrm{U} atoms to { }^{206} \mathrm{~Pb} equal to 0.5 . Then the age of the rock sample is:\left[T_{1 / 2}(238 U)=4.5 \times 10^9 \mathrm{yr}\right]
 

Option: 1

7.1 \times 10^9 \mathrm{yrs}


Option: 2

5.1 \times 10^6 \mathrm{yrs}


Option: 3

7.1 \times 10^8 \mathrm{yrs}


Option: 4

6.1 \times 10^5 \mathrm{yrs}


Answers (1)

best_answer

Let \mathrm{N}_{\mathrm{o}} be the initial number uranium atom \left({ }^{238} \mathrm{U}\right), then the number of { }^{238} \mathrm{U}, after time \mathrm{t} is given by

         \mathrm{NU}=\mathrm{N}_0 \mathrm{e}^{-\lambda t}, while the number of { }^{206} \mathrm{~Pb}atoms is given by \left(\mathrm{N}_0-\mathrm{N}\right) assuming that the intermediate states are short-lived.

\mathrm{Np}_0=\mathrm{N}_0-\mathrm{N}=\mathrm{N}_0\left(1-\mathrm{e}^{-\lambda t}\right), 

where \lambda=0.693 / \mathrm{T}_{1 / 2}, \mathrm{~T}_{1 / 2}=4.5 \times 10^9 \mathrm{yr},  the half life of \operatorname{uranium}\left({ }^{28} \mathrm{U}\right)
The ratio, \quad \mathrm{R}=\frac{N_o e^{-\lambda t}}{N_o\left(1-e^{-\lambda t}\right)} ; t=\frac{1}{\lambda}\left(1+\frac{1}{R}\right)

Substituting the given values of R, we get,

\mathrm{T}=7.1 \times 10^9 \mathrm{yr}

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Delhi’s CSE, EE most opted; last 2 year’s closing ranks for all categories
anam-khan

Must provide CBSE 11th registration numbers to JEE Mains aspirants, board tells schools
anam-khan

JEE Main 2026: Engineering at NIT Surathkal tougher for insiders, show BTech cut-offs

Latest Question