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Chapter 38: Problem 48

Show that the decay constant and half-life are related by \(t_{1 / 2}=\ln 2 / \lambda \simeq 0.693 / \lambda.\)

Short Answer

Expert verified
The decay constant (\( \lambda \)) and the half-life (\( t_{1/2} \)) of a radioactive material are related by the equation \( t_{1 / 2}=\ln 2 / \lambda \simeq 0.693 / \lambda\). This relationship is derived based on the mathematical description of radioactive decay, \( N = N_0 e^{-\lambda t} \), where \( N \) is the amount of substance at time \( t \), \( N_0 \) is the initial amount, and \( e \) is the base of the natural logarithm.

Step by step solution

01

Define half-life and decay constant

Start by defining the terms in the equation. The decay constant (\( \lambda \)) is the probability per unit time that a nucleus will decay, and half-life (\( t_{1/2} \)) is the time required for half of the radioactive nuclei in a sample to decay.
02

Derive the equation for radioactive decay

The mathematical description of radioactive decay is given by the equation \( N = N_0 e^{-\lambda t} \), where \( N \) is the amount of radioactive substance at time \( t \), \( N_0 \) is the initial amount, and \( e \) is the base of the natural logarithm. For half-life, the amount of substance \( N \) should be half of the initial amount \( N_0 \). So we substitute \( N = \frac{N_0}{2} \) into the equation and get \( \frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \).
03

Solve for \( t_{1/2} \)

Divide both sides by \( N_0 \), and then take natural logarithms on both sides. This gives you \( \ln(1/2) = -\lambda t_{1/2} \). Now, it’s just a matter of simplifying to solve for \( t_{1/2} \): \( t_{1/2} = \frac{\ln(1/2)}{-\lambda} \). The value of \( \ln(1/2) \) is approximately -0.693, hence the equation becomes \( t_{1/2} = \frac{0.693}{\lambda} \).

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