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Chapter 19: Problem 23

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

Short Answer

Expert verified
The half-life (T₁/₂) of the nuclide can be calculated using the relation \(T_{1/2} = \frac{ln(2)}{k}\), where k is the given rate constant \(k = 1.0 \times 10^{-3} \mathrm{h}^{-1}\). Plugging in the given value for k, we find that the half-life of the nuclide is approximately \(T_{1/2} \approx 693.1 \mathrm{h}\).

Step by step solution

01

Note down the given values

We are given the rate constant (k) for a radioactive nuclide: \(k = 1.0 \times 10^{-3} \mathrm{h}^{-1}\) Our task is to find the half-life (T₁/₂) of the nuclide.
02

Use the formula to find the half-life

Now, we will use the formula that relates the rate constant (k) with half-life (T₁/₂): \(T_{1/2} = \frac{ln(2)}{k}\) Here, ln(2) is the natural logarithm of 2. Plug in the given value for k and find the half-life.
03

Calculate the half-life

Insert the given value for k into the formula: \(T_{1/2} = \frac{ln(2)}{1.0 \times 10^{-3} \mathrm{h}^{-1}}\) Calculate the half-life: \(T_{1/2} \approx \frac{0.6931}{1.0 \times 10^{-3} \mathrm{h}^{-1}} = 693.1 \mathrm{h}\) So, the half-life of the nuclide is approximately 693.1 hours.

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Most popular questions from this chapter

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{68} \mathrm{Ga}\) (electron capture) c. \({ }^{212} \mathrm{Fr}(\alpha)\) b. \(^{62} \mathrm{Cu}\) (positron) d. \({ }^{129} \mathrm{Sb}(\beta)\)

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