Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Uranium ore on the earth at the present time topically have a composition consisting of $99.3 \%$ of the isotope ${ }_{92} U^{238}$ and $0.7 \%$ of the isotope ${ }_{92} U^{235}$ . The half lives of these isotopes are $4.47 \times 10^9 \mathrm{y}$ and $7.04 \times 10^8 \mathrm{y}$ respectively. If these isotopes were equally abundant when the earth was formed. Estimate the age of the earth.

$\text { Given } \ln =\left(\frac{993}{7}\right)=4.95$
Option: 1
$5.9 \times 10^9 \mathrm{y}$

Option: 2
$6.9 \times 10^9 \mathrm{y}$

Option: 3
$7.9 \times 10^9 \mathrm{y}$

Option: 4
$8.9 \times 10^9 \mathrm{y}$

Answers (1)

best_answer

$\lambda_1=\frac{0.693}{4.47 \times 10^9} \mathrm{y}^{-1} \quad, \quad \lambda_2=\frac{0.693}{7.04 \times 10^8} \mathrm{y}^{-1}$

$\mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-\lambda_1 \mathrm{t}}\, \, , \quad \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda_2 \ \mathrm{t}}$

$\Rightarrow \frac{N_2}{N_1}=\mathrm{e}(\lambda_1-\lambda_2) \mathrm{t}=\frac{0.7}{99.3}$

$\Rightarrow \quad\left(\lambda_2-\lambda_1\right) \mathrm{t}=\ln \left(\frac{993}{7}\right)$
$\Rightarrow \quad \mathrm{t}=\frac{1}{\frac{0.693}{7.04 \times 10^8}-\frac{0.693}{4.47 \times 10^9}}=\ln \left(\frac{993}{7}\right)$

$=5.9 \times 10^9 \mathrm{yr}$

Posted by

manish

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2026 Tie-Breaking Rule Explained: Here’s how ranks are decided

anam-khan

NEET 2026: Odisha aspirant dies by suicide fearing failure

anam-khan

NEET 2026 Answer Key (Out) LIVE: NTA to open challenge window shortly; direct link, result date, counselling

Latest Question