$(a)\;9 \times10^{9} years \qquad(b)\;6.3 \times10^{9} years \qquad(c)\;3.78\times10^{9} years \qquad(d)\;None$

Answer : (c) $\;3.78 \times10^{9} years $

Explanation :

Let initial no.of U- atoms = $\;N_{0}$

After time t , (age of rock ) , Let no.of atoms remaining decayed = N

$\large\frac{238 N}{26 (N_{0}-N)}=\large\frac{4}{3}$

$\large\frac{N_{0}}{N}=1.79$

$t=\large\frac{\large\frac{T log N_{0}}{N}}{log 2}=\large\frac{4.5 \times10^{9}\times log 1.79}{0.301}$

$=3.78\times 10^{9} years $

Ask Question

Tag:MathPhyChemBioOther

Take Test

...