The expression for calculating half-life for a radioactive decay is:

\(t_{1/2} = \frac{0.693}{K}\)

The correct answer is option 3. \(t_{1/2} = \frac{0.693}{K}\).

The half-life (\(t_{1/2}\)) of a radioactive substance is the time it takes for half of the radioactive nuclei in a sample to decay. This means that after one half-life, only half of the original amount of the radioactive substance remains.

Radioactive decay follows first-order kinetics, meaning the rate of decay is directly proportional to the number of undecayed nuclei. The mathematical expression for radioactive decay is given by:

\(N = N_0 e^{-Kt}\)

\(N\): The number of undecayed nuclei at time \(t\).

\(N_0\): The initial number of undecayed nuclei.

\(K\): The decay constant, which is specific to each radioactive substance.

\(t\): The time elapsed.

To find the half-life, we set \(N\) equal to half of \(N_0\) (since after one half-life, only half of the original substance remains):

\(\frac{N_0}{2} = N_0 e^{-Kt_{1/2}}\)

Now, let's solve for \(t_{1/2}\):

Dividing both sides by \(N_0\):

\(\frac{1}{2} = e^{-Kt_{1/2}}\)

Taking log on both sides:

\(\ln\left(\frac{1}{2}\right) = \ln\left(e^{-Kt_{1/2}}\right)\)

Simplyfying using the logarithm property \(\ln(e^x) = x\):

\(\ln\left(\frac{1}{2}\right) = -Kt_{1/2}\)

We know, \(\ln\left(\frac{1}{2}\right)\) is equal to \(-\ln(2)\), thus

\(-\ln(2) = -Kt_{1/2}\)

Dividing both sides by \(-K\) to solve for \(t_{1/2}\):

\(t_{1/2} = \frac{\ln(2)}{K}\)

Substituting \(\ln(2)\) with its numerical value \(0.693\):

\(t_{1/2} = \frac{0.693}{K}\)

The decay constant (\(K\)) represents how quickly a substance decays. A larger \(K\) means the substance decays faster, leading to a shorter half-life.The factor 0.693 (which is \(\ln(2)\)) comes from the natural logarithm of 2, reflecting the mathematical nature of exponential decay. The half-life \(t_{1/2}\) is inversely proportional to the decay constant \(K\). The correct expression for calculating the half-life is:

\(t_{1/2} = \frac{0.693}{K}\)