72) The number of half-lives (T1/2) 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
$\frac{N}{N_{0}}=e^{-\lambda
t}$
$0.03=e^{-\lambda t}$
$\ln(0.03)=-\lambda t$
$\lambda t=3.506$
$\frac{0.693}{t_{1/2}}
\times t=3.506$
$t=\frac{3.506}{0.693}t_{1/2}$
$t=5.059t_{1/2}$
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