Calculation of true thickness of the bed

 

72) The number of half-lives (T_1/2) required for a certain amount of radioactive isotopes in a rock to reduce to 3% of its original amount is_____

 

(Thanks to Chandrasekhar, ANU)

Solution:

 

We know that the $\frac{N}{N_{0}}=e^{-\lambda t}$---------------(1)

 

  Given that $\frac{N}{N_{0}}$=3% = 0.03

 

 By substituting the above value in equation (1) w get

 

$\frac{N}{N_{0}}=e^{-\lambda t}$

 

$0.03=e^{-\lambda t}$

 

$\ln(0.03)=-\lambda t$

 

$\lambda t=3.506$

 

$\frac{0.693}{t_{1/2}} \times t=3.506$

 

$t=\frac{3.506}{0.693}t_{1/2}$

 

$t=5.059t_{1/2}$