If the half-life period of a radioactive isotope is 10 s, then its average life will be:

14.4 s

The correct answer is option 1. 14.4 s.

Half-Life (\(t_{1/2}\)):

The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. It’s a measure of the rate of decay but doesn’t directly tell us how long an individual atom will live on average.

Average Life (Mean Life, \(\tau\)):

The average life of an isotope is the average time an atom of the isotope will exist before decaying. It’s a measure that helps us understand the expected lifespan of an atom in a probabilistic sense.

Relationship Between Half-Life and Average Life:

The average life (\(\tau\)) and the half-life (\(t_{1/2}\)) are related through the natural logarithm of 2. This relationship arises from the exponential nature of radioactive decay.

The decay of a radioactive substance follows an exponential function:
\(N(t) = N_0 e^{-\lambda t}\)

Where:

\(N(t)\) is the number of radioactive atoms at time \(t\).

\(N_0\) is the initial number of radioactive atoms.

\(\lambda\) is the decay constant, which is related to the half-life.

The half-life (\(t_{1/2}\)) is related to the decay constant (\(\lambda\)) by:

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

Rearranging this to solve for \(\lambda\):

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

The average life (\(\tau\)) is the reciprocal of the decay constant:

\(\tau = \frac{1}{\lambda}\)

Substitute \(\lambda\) from the above relationship:

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

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

Given:

Half-life (\(t_{1/2}\)) = 10 seconds

The average life (\(\tau\)):

\(\tau = \frac{t_{1/2}}{\ln 2} = \frac{10}{0.693} \approx 14.4 \text{ seconds}\)

Summary:

Half-life tells us how long it takes for half of the radioactive atoms to decay. Average life gives us the expected time an atom will exist before decaying, calculated as \( \frac{t_{1/2}}{\ln 2} \).

So, with a half-life of 10 seconds, the average life of the radioactive isotope is approximately 14.4 seconds.