$(a)\;2.12 \times 10^{13} years\qquad(b)\;1.26 \times 10^{7} years\qquad(c)\;1.87 \times 10^{9} years\qquad(d)\;3.2 \times 10^{10} years$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : (c) $\;1.87 \times 10^{9} years$

Explanation :

In this problem it is assumed that half life of intermediate products is very small in comparision to that of $\;U^{238}$

Let required age is 't' . If N be initial no. of $\;U^{238}\;$ nuclei then

$\large\frac{N_{0}e^{-\lambda t}}{N_{0}\; (1-e^{\lambda t})}=3$

$4 e^{-\lambda t}=3 \qquad \; \lambda t=ln (\large\frac{4}{3})$

$t=\large\frac{ln(\large\frac{4}{3})}{\lambda}=\large\frac{ln(\large\frac{4}{3})}{ln 2}\;T$

$=1.87 \times 10^{9} years \;.$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...