The average kinetic energy of an atom in a monatomic ideal gas is given by KE=3/2kT, where k=1.38*10^-23J/K and T is the Kelvin temperature of the gas. Determine the de Broglie wavelength of a helium atom (mass=6.65*10^-27kg) that has the average kinetic energy at room temperature (293K).
I don't get this question at all and I have no idea how to do it. I NEED HELP AND PLEASE EXPLAIN!THANKS A LOT!
To determine the de Broglie wavelength of a helium atom with the average kinetic energy at room temperature, you can follow these steps:
Step 1: Calculate the average kinetic energy using the given formula KE = 3/2kT.
- Given: k = 1.38 * 10^-23 J/K and T = 293 K
- Substitute the values and calculate: KE = (3/2) * (1.38 * 10^-23 J/K) * 293 K
Step 2: Use the kinetic energy equation to find the velocity of the helium atom.
- The kinetic energy equation is given by KE = (1/2)mv^2
- Rearrange the equation to solve for velocity: v = sqrt((2 * KE) / m)
- Substitute the calculated kinetic energy and the mass of helium (6.65 * 10^-27 kg) into the equation, and calculate the velocity.
Step 3: Calculate the de Broglie wavelength using the formula λ = h / p.
- The de Broglie wavelength is equivalent to Planck's constant (h) divided by the momentum of the particle (p).
- Planck's constant is denoted as h = 6.62607015 × 10^-34 J·s.
- Calculate the momentum, p, which is given by p = m * v, where m is the mass and v is the velocity.
- Substitute the calculated values of h and p into the equation to find the de Broglie wavelength.
By following these steps, you will be able to determine the de Broglie wavelength of the helium atom with the average kinetic energy at room temperature.