The rate constant for a certain radioactive nuclide is \(1.0 \mathrm{X}\) \(10^{-3} \mathrm{~h}^{-1}\). What is the half-life of this nuclide?

Short Answer

Expert verified
The half-life of the radioactive nuclide with a rate constant of \(1.0 \times 10^{-3} \text{ h}^{-1}\) can be calculated using the formula \(t_{1/2} = \frac{\ln{2}}{k}\). Substituting the given values, the half-life is approximately 693.1 hours.

Step by step solution

01

Identifying the given values

The rate constant (k) for the radioactive nuclide is given as \(1.0 \times 10^{-3} \text{ h}^{-1}\). Now, we need to find the half-life (t½) of this nuclide.
02

Write down the formula for half-life related to rate constant

The formula that relates the half-life (t½) to the rate constant (k) is as follows: \[ t_{1/2} = \frac{\ln{2}}{k} \]
03

Substitute the given values into the formula

Now, we will substitute the given value of k into the formula to calculate the half-life: \[ t_{1/2} = \frac{\ln{2}}{(1.0 \times 10^{-3}\text{ h}^{-1})} \]
04

Calculate the half-life of the nuclide

Using a calculator or logarithm table, we can find the value of log2: \[ \ln{2} \approx 0.6931 \] Now, substitute this value into the formula and calculate t½: \[ t_{1/2} = \frac{0.6931}{(1.0 \times 10^{-3}\text{ h}^{-1})} \approx 693.1 \text{ h} \]
05

Write down the final result

The half-life of this radioactive nuclide is approximately 693.1 hours.

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