What is the half-life of an isotope if 436 seconds are required for the number of cpm to diminish from 14,227 to 3672 Show your calculations.
The correct answer and explanation is :
To determine the half-life (( t_{1/2} )) of the isotope, we use the first-order decay formula:
[
N = N_0 e^{-\lambda t}
]
where:
- ( N_0 ) is the initial count rate (14,227 cpm),
- ( N ) is the final count rate (3,672 cpm),
- ( \lambda ) is the decay constant,
- ( t ) is the time elapsed (436 seconds),
- ( e ) is the natural logarithm base (~2.718).
Step 1: Solve for ( \lambda )
Taking the natural logarithm on both sides:
[
\ln \left( \frac{N}{N_0} \right) = -\lambda t
]
Substituting the values:
[
\ln \left( \frac{3672}{14227} \right) = -\lambda (436)
]
[
\ln (0.2581) = -\lambda (436)
]
[
-1.354 = -\lambda (436)
]
[
\lambda = \frac{1.354}{436}
]
[
\lambda = 0.00311 \text{ s}^{-1}
]
Step 2: Determine Half-Life
The relationship between the half-life and decay constant is:
[
t_{1/2} = \frac{\ln 2}{\lambda}
]
[
t_{1/2} = \frac{0.693}{0.00311}
]
[
t_{1/2} = 223 \text{ seconds}
]
Explanation (300 Words)
The half-life of an isotope is the time required for half of a radioactive substance to decay. It is a key property in nuclear chemistry and radiometric dating.
In this problem, we determined the half-life using first-order decay kinetics. Radioactive decay follows the exponential law, meaning the quantity of a substance decreases at a rate proportional to its current amount. This is described by the equation ( N = N_0 e^{-\lambda t} ), where ( \lambda ) is the decay constant.
By rearranging and solving for ( \lambda ), we found it to be 0.00311 s(^{-1}). The half-life is calculated using the relationship ( t_{1/2} = \frac{\ln 2}{\lambda} ), which gives 223 seconds.
This means that every 223 seconds, half of the radioactive atoms decay. Understanding half-life is crucial in fields such as medicine (radiotherapy), archaeology (carbon dating), and nuclear power.