Question

starting with a sample of Pure \({ }^{66} \mathrm{cu},(7 / 8)\) of it decays into \(\mathrm{Zn}\), 15 minutes. The half life the sample is (A) \(5 \mathrm{~min}\) (B) \(7.5 \mathrm{~min}\) (C) \(10 \mathrm{~min}\) (D) \(15 \mathrm{~min}\)

Short answer

The half-life of the given sample of Cu-66 is 5 minutes, as determined by using the decay formula and solving for the half-life. The correct option is (A) \(5 \mathrm{~min}\).

Step-by-step solution

Identify the known values

We are given the following values: - Initial amount of Cu-66: \(N_0\) - Amount of Cu-66 remaining after 15 minutes: \((1/8) N_0\), because 7/8 of it decays, so \((1 - (7/8)) N_0 = (1/8) N_0\). - Time elapsed (t): 15 minutes

Plug the known values into the decay formula

Now we will plug these values into the decay formula: \[(1/8) N_0 = N_0 (1/2)^{15/T}\]

Solve for the half-life (T)

First, we will cancel \(N_0\) from both sides of the equation, since we are only interested in the half-life. \[(1/8) = (1/2)^{15/T}\] To solve for \(T\), we can take the logarithm of both sides, using the base \(1/2\). So we have: \[\log_{1/2} (1/8) = \frac{15}{T}\] The logarithm on the left side can be simplified: \[\log_{1/2} (1/2)^3 = \frac{15}{T}\] Which gives us: \[3 = \frac{15}{T}\]

Determine the half-life

Now we can solve for \(T\) by multiplying both sides by \(T\) and dividing both sides by 3: \[T = \frac{15}{3}\] \[T = 5\] So, the half-life of the given sample is 5 minutes. The correct option is: (A) \(5 \mathrm{~min}\)