Question

If a substance decays from 32 g to 1 g in 25 days, then its half-life is:

• A. 3 days
• B. 4 days
• C. 5 days
• D. 6 days

Hint:

The decay constant \(k\) can be calculated from the exponential decay equation, and the half-life can be derived using the formula \(T_{1/2} = \frac{\ln(2)}{k}\).

Answer 1

The Correct Option is C

The decay of a substance follows the exponential decay law: \[ N(t) = N_0 e^{-kt} \] Where \(N(t)\) is the amount of substance remaining at time \(t\), \(N_0\) is the initial amount, \(k\) is the decay constant, and \(t\) is the time.
We are given that the substance decays from 32 g to 1 g in 25 days. This means that: \[ \frac{N(t)}{N_0} = \frac{1}{32} \] Using the formula for exponential decay, we can solve for \(k\): \[ \frac{1}{32} = e^{-k \cdot 25} \] Taking the natural logarithm of both sides: \[ \ln\left(\frac{1}{32}\right) = -k \cdot 25 \] \[ k = \frac{\ln(32)}{25} \] Now, the half-life \(T_{1/2}\) is related to \(k\) by: \[ T_{1/2} = \frac{\ln(2)}{k} \] Substituting the value of \(k\) from above: \[ T_{1/2} = \frac{\ln(2)}{\frac{\ln(32)}{25}} = \frac{25 \ln(2)}{\ln(32)} \approx 5 \, \text{days} \] Thus, the half-life of the substance is 5 days.