Question

Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use `u_(avg) = sqrt((8k_BT)/(pi m))`

Answer 1

I got `"0.0794 nm"`, or `"79.4 pm"`, over twice its atomic radius.

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Well, assuming that the `"He"` atoms are all moving in the same direction, and follow a Maxwell-Boltzmann distribution:

`f(u) = 4pi (m/(2pik_BT))^(3//2) u^2 e^(-m u^2//2k_BT)`

they thus have the average speed

`u_(avg) = int_(0)^(oo) uf(u)du = [ . . . ] = sqrt((8k_B T)/(pim))`,

where:

• `k_B = 1.38065 xx 10^(-23)` is the Boltzmann constant in `"J/K"`, or `"kg"cdot"m"^2"/s"^2cdot"K"`.
• `T` is the temperature in `"K"`.
• `m` is the per-particle mass in `"kg"`, i.e. `M//N_A = m`, where `M` is the molar mass in `"kg/mol"` and `N_A = 6.0221413 xx 10^(23) "mol"^(-1)`.

We then use this average speed as the velocity `v` in the de Broglie relation

`lambda = h/(mv)`,

where `h = 6.626 xx 10^(-34) "J"cdot"s"`, or `"kg"cdot"m"^2"/s"`, is Planck's constant and `m` is as defined before,

since the particles are assumed to all be moving in the same direction so that the velocity is the speed.

Therefore, the wavelength is:

`color(blue)(lambda) = h/m xx sqrt((pim)/(8k_B T))`

`= sqrt((pih^2)/(8mk_B T))`

`= sqrt((pi(6.626 xx 10^(-34) cancel("kg")cdot"m"^cancel(2)"/"cancel("s"))^2)/(8(0.0040026 cancel("kg")"/"cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23)))(1.38065 xx 10^(-23) cancel("kg")cdotcancel("m"^2)"/"cancel("s"^2)cdotcancel"K")(298.15 cancel("K"))))`

`= 7.938 xx 10^(-11)` `"m"`

`=` `color(blue)("0.0794 nm")`,

which is in the gamma-ray region of the EM spectrum.