Problem 69 - Chemistry The Science in Context

Question

Kinetic molecular theory tells us that helium atoms at \(500 \mathrm{K}\) are in constant random motion. a. Calculate the root-mean-square speed of helium atoms at \(500 \mathrm{K}\) b. What is the wavelength of a helium atom at \(500 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Question: Calculate the root-mean-square speed of helium atoms at 500 K and their corresponding wavelength. Answer: The root-mean-square speed of helium atoms at 500 K is approximately 1372 m/s, and their corresponding wavelength is approximately 7.243 x 10^(-11) m.

Step 1: a. Calculate the root-mean-square speed of helium atoms at \(500 \mathrm{K}\)

First, we need to determine the mass of a helium atom. The molar mass of helium is \(4.0026 \frac{\mathrm{g}}{\mathrm{mol}}\). We can convert this to the mass of a single atom by dividing by Avogadro's number (\(N_{\rm A}\)) and converting grams to kilograms: \( m_\text{He}=\frac{4.0026 \frac{\mathrm{g}}{\mathrm{mol}}}{6.022\times 10^{23}{\rm mol^{-1}}}\cdot\frac{1\mathrm{kg}}{1000\mathrm{g}}=6.646\times10^{-27}\mathrm{kg}\) Now, we will use the formula for the root-mean-square speed: \(v_\mathrm{rms} = \sqrt{\frac{3kT}{m}}\) We will plug in the values: \(k=1.38 \times 10^{-23} \mathrm{J\ K^{-1}}, T=500\mathrm{K}, \mathrm{and} \ m=6.646\times10^{-27}\mathrm{kg}\) \(v_\mathrm{rms} = \sqrt{\frac{3(1.38 \times 10^{-23} \mathrm{J\ K^{-1}})(500 \mathrm{K})}{6.646\times10^{-27}\mathrm{kg}}}=1371.99 \frac{\mathrm{m}}{\mathrm{s}}\) So, the root-mean-square speed of helium atoms at \(500 \mathrm{K}\) is approximately \(1372 \frac{\mathrm{m}}{\mathrm{s}}\).

Step 2: b. What is the wavelength of a helium atom at \(500 \mathrm{K} ?\)

To find the wavelength of a helium atom at \(500 \mathrm{K}\), we first need to determine its momentum. The momentum of a particle is given by: \(p = mv\) where m is the mass and v is the velocity of the particle. Since we are considering the root-mean-square speed, we will use the value of \(v=\frac{mv_\mathrm{rms}}{m}\) Next, we will use de Broglie's formula to find the wavelength: \(\lambda = \frac{h}{p}\), where \(h=6.626\times10^{-34}\mathrm{Js}\) is Planck's constant. We can plug in the values of h, m, and \(v_\mathrm{rms}\) we found earlier: \(\lambda = \frac{6.626\times10^{-34}\mathrm{Js}}{6.646\times10^{-27}\mathrm{kg}\cdot 1371.99 \frac{\mathrm{m}}{\mathrm{s }}}=7.243\times 10^{-11}\mathrm{m}\) So, the wavelength of a helium atom at \(500 \mathrm{K}\) is approximately \(7.243 \times 10^{-11}\mathrm{m}\).

Key Concepts

Root-mean-square speed

The concept of root-mean-square speed (\(v_{\text{rms}}\)) is quite fascinating and essential to understand the motion of particles, especially gases, as described by the kinetic molecular theory. By definition, the root-mean-square speed is the square root of the average squared speeds of all the molecules in a gas.The formula to calculate the root-mean-square speed is:\[v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\]Here,

• \(k\)is the Boltzmann constant, equal to \(1.38 \times 10^{-23} \text{J}\ \text{K}^{-1}\).
• \(T\)is the temperature of the gas in Kelvin (\(K\)).
• \(m\)is the mass of a single molecule in kilograms \(\text{kg}\).

Calculating the root-mean-square speed is useful because it gives us a sense of how fast particles are moving on average at a given temperature. In the exercise on helium atoms, we calculated this to be \(1372 \text{m}/\text{s}\) at \(500 \text{K}\), indicating that helium atoms are moving swiftly at this temperature.

de Broglie wavelength

The concept of de Broglie wavelength brings a touch of quantum mechanics into the classical physics world. It deals with the wave-particle duality of matter. In essence, any particle with momentum possesses a wavelength, known as the de Broglie wavelength. The de Broglie relation is given by:\[\lambda = \frac{h}{p}\]Where,

• \(\lambda\) is the wavelength of the particle.
• \(h\) is Planck's constant (\(6.626\times10^{-34} \text{J}\text{s}\)).
• \(p\) is the momentum of the particle, calculated as the product of mass \(m\) and velocity \(v\), that is \(p = mv\).

When calculating the de Broglie wavelength for helium atoms at \(500 \text{K}\), we first determine their momentum using the root-mean-square speed. Subsequently, inserting these values into the de Broglie equation, we find that each helium atom exhibits a very tiny wavelength of approximately \(7.243 \times 10^{-11} \text{m}\). This minuscule wavelength is evidence of the particle-like behavior dominant at macroscopic scales, with wave properties overshadowed by other factors at this level.

Helium atom mass

The mass of a helium atom is a fundamental piece of information for many calculations, particularly when dealing with kinetic energy or velocity in gaseous states like helium. To find the individual mass of a helium atom, we start from its molar mass.The molar mass of helium is \(4.0026 \text{g/mol}\). To convert this to the mass of a single atom, we need Avogadro's number (\(N_{\rm A} = 6.022 \times 10^{23} \text{mol}^{-1}\)), which gives us the number of atoms in a mole. The calculation becomes:\[m_{\text{He}}=\frac{4.0026 \times 10^{-3} \text{kg/mol}}{6.022 \times 10^{23} \text{mol}^{-1}}\] After simplification, we find that the mass of one helium atom is approximately \(6.646 \times 10^{-27} \text{kg}\). This incredibly small mass highlights the microscopic nature of atoms, requiring large numbers to measure measurable quantities, like grams or moles. Understanding this atomic-scale mass is crucial when performing calculations concerning speed, energy, or momentum on singular atoms in kinetic molecular realm.