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Chapter 18: Problem 69

Using the kinetic molecular theory (see Section 8.6), calculate the root mean square velocity and the average kinetic energy of \(_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{K}\). (See Exercise 50 for the appropriate mass values.)

Short Answer

Expert verified
The root-mean-square velocity of \(_1^2\mathrm{H}\) nuclei at the given temperature is approximately \(2.2345×10^5\: \mathrm{m/s}\), and the average kinetic energy is around \(8.2936×10^{-16}\: \mathrm{J}\).

Step by step solution

01

Find the mass of \(_1^2\mathrm{H}\) nucleus

The mass of \(_1^2\mathrm{H}\) nucleus can be found by converting its atomic mass unit (u) to kilograms (kg). The atomic mass of \(_1^2\mathrm{H}\) is approximately 2 u. The conversion factor between atomic mass units and kilograms is 1 u = 1.66054 x 10^{-27} kg. So, the mass of a single \(_1^2\mathrm{H}\) nucleus is: \[m = 2\: \mathrm{u} \times \frac{1.66054×10^{-27} \: \mathrm{kg}}{1\: \mathrm{u}} = 3.32108×10^{-27} \: \mathrm{kg}\]
02

Calculate the root-mean-square velocity

Now that we have the mass of a single \(_1^2\mathrm{H}\) nucleus, we can use the root-mean-square velocity formula: \[v_{rms} = \sqrt{\frac{3kT}{m}}\] Here, \(k\) is the Boltzmann constant, which is equal to 1.380649 x 10^{-23} J/K. The temperature \(T\) is given as 4 x 10^7 K. \[v_{rms} = \sqrt{\frac{3 (1.380649×10^{-23}\: \mathrm{J/K})(4×10^{7}\: \mathrm{K})}{3.32108×10^{-27} \: \mathrm{kg}}}\] \[v_{rms} = \sqrt{\frac{1.65857×10^{-15}\: \mathrm{J}}{3.32108×10^{-27} \: \mathrm{kg}}}\] \[v_{rms} \approx 2.2345×10^5\: \mathrm{m/s}\] The root-mean-square velocity of \(_1^2\mathrm{H}\) nuclei at the given temperature is around 2.2345 x 10^5 m/s.
03

Calculate the average kinetic energy

Using the root-mean-square velocity found in the previous step, we can calculate the average kinetic energy using the following formula: \(\overline{KE} = \frac{1}{2}m(v_{rms})^2\) \(\overline{KE} = \frac{1}{2}(3.32108×10^{-27}\: \mathrm{kg})(2.2345×10^5\: \mathrm{m/s})^2\) \(\overline{KE} \approx 8.2936×10^{-16}\: \mathrm{J}\) The average kinetic energy of \(_1^2\mathrm{H}\) nuclei at the given temperature is around 8.2936 x 10^{-16} J.

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