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Chapter 19: Problem 169

The half-life of a radioactive substance is \(20 \mathrm{~min}\). The approximate time interval \(\left(t_{2}-t_{1}\right)\) between the time \(t_{2}\) when \(\frac{2}{3}\) of it has decayed and time \(t_{1}\) when \(\frac{1}{3}\) of it had decayed is (A) \(7 \mathrm{~min}\) (B) \(14 \mathrm{~min}\) (C) \(20 \mathrm{~min}\) (D) \(28 \mathrm{~min}\)

Short Answer

Expert verified
The time interval between the time \(t_{2}\) when \(\frac{2}{3}\) of the radioactive substance has decayed and time \(t_{1}\) when \(\frac{1}{3}\) of it has decayed is \(20 \mathrm{~min}\).

Step by step solution

01

Write the radioactive decay formula

Radioactive decay can be described using the following formula: \[A(t) = A_{0} \cdot 2^{-\frac{t}{h}}\] where \(A_0\) is the initial amount of the substance, \(A(t)\) is the remaining amount of the substance at time \(t\), \(h\) is the half-life of the substance, and \(t\) is the time.
02

Solve for A(t) for \(\frac{1}{3}\) and \(\frac{2}{3}\) decayed

From our problem, we know that at time \(t_1\), \(\frac{1}{3}\) of the substance has decayed. Therefore, the remaining amount is \(\frac{2}{3}\) of the initial amount: \[A(t_1) = \frac{2}{3}A_0\] Similarly, at time \(t_2\), \(\frac{2}{3}\) of the substance has decayed, leaving us with \[\frac{1}{3}A_0\].
03

Substitute A(t) into the decay formula for both time points:

Next, substitute the expressions for \(A(t_1)\) and \(A(t_2)\) into the decay formula and solve for \(t_1\) and \(t_2\). For time \(t_1\): \[\frac{2}{3}A_0 = A_0 \cdot 2^{-\frac{t_1}{20}}\] For time \(t_2\): \[\frac{1}{3}A_0 = A_0 \cdot 2^{-\frac{t_2}{20}}\]
04

Solve for \(t_1\) and \(t_2\) by eliminating the \(A_0\) term and using logarithms

Divide the first equation by the second equation to eliminate the initial amount \(A_0\) term: \[\frac{\frac{2}{3}A_0}{\frac{1}{3}A_0} = \frac{2^{-\frac{t_1}{20}}}{2^{-\frac{t_2}{20}}}\] This simplifies to: \[2 = 2^{\frac{t_2-t_1}{20}}\] Now take the logarithm (base 2) of both sides: \[\log_2 2 = \log_2 {2^{\frac{t_2-t_1}{20}}}\]
05

Solve for \(t_2 - t_1\)

Using the logarithm properties, we can simplify the equation to: \[1 = \frac{t_2-t_1}{20}\] Solving for \(t_2 - t_1\), we get: \[t_2-t_1 = 20 \mathrm{~min}\] Therefore, the correct answer is (C) \(20 \mathrm{~min}\).

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