GATE-2018 (24) :Wyllie – Time average equation

21) In an unweathered sample of detrital Sedimentary rock. The ratio of an unstable isotope to its stable daughter isotope is 1/15. If no daughter were present at the time the rock cooled, and the half-life of the radioactive parent isotope is 50Ma, how old is the rock?

(Thank to Neeraja,AU)

Solution:

Formula :

$N=N_o(\frac{1}{2})^{\frac{t}{t_{1/2}}}---(1)$

N – unstable isotope

N_o – stable isotope

t – time

t_1/2 – half life

From above solution,

$\frac{N}{N_o}=\frac{1}{15}$

$t_{1/2}=50Ma$

t=?

From eq.(1)

$\frac{N}{N_o}=(\frac{1}{2})^{\frac{t}{t_{1/2}}}$

$\frac{1}{15}=(\frac{1}{2})^{\frac{t}{50\times 10^6}}$

$log(\frac{1}{15})=log(\frac{1}{2})^{\frac{t}{50\times 10^6}}$

$-1.176 =\frac{t}{50\times 10^6}log(\frac{1}{2})$

$-1.176 =\frac{t}{50\times 10^6}\times (-0.30)$

$\frac{t}{50\times10^6}=\frac{1.176}{0.30}$

$\frac{t}{50\times10^6}=3.92$

$t=3.92\times50\times10^6$

t=196 Ma

The age of the rock is 196 Ma

Reference: Geo Knowledge, MCQs series