Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use $u_{a v g} = \sqrt{\frac{8 k_{B} T}{π m}}$ $u_(avg) = sqrt((8k_BT)/(pi m))$

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Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use $u_{a v g} = \sqrt{\frac{8 k_{B} T}{π m}}$ $u_(avg) = sqrt((8k_BT)/(pi m))$

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Answer 1 · 2017-07-18T19:25:48

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

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

$f (u) = 4 π {(\frac{m}{2 π k_{B} T})}^{3 / 2} u^{2} e_{B}^{- m u^{2} / 2 k_{B} T}$ $f(u) = 4pi (m/(2pik_BT))^(3//2) u^2 e^(-m u^2//2k_BT)$

they thus have the average speed

$u_{a v g} = \int_{0}^{\infty} u f (u) d u = [. . .] = \sqrt{\frac{8 k_{B} T}{π m}}$ $u_(avg) = int_(0)^(oo) uf(u)du = [ . . . ] = sqrt((8k_B T)/(pim))$ ,

where:

$k_{B} = 1.38065 \times 10^{- 23}$ $k_B = 1.38065 xx 10^(-23)$ is the Boltzmann constant in $J/K$ $"J/K"$ , or $kg \cdot m^{2} {/s}^{2} \cdot K$ $"kg"cdot"m"^2"/s"^2cdot"K"$ .

$T$ $T$ is the temperature in $K$ $"K"$ .

$m$ $m$ is the per-particle mass in $kg$ $"kg"$ , i.e. $M / N_{A} = m$ $M//N_A = m$ , where $M$ $M$ is the molar mass in $kg/mol$ $"kg/mol"$ and $N_{A} = 6.0221413 \times 10^{23} {mol}^{- 1}$ $N_A = 6.0221413 xx 10^(23) "mol"^(-1)$ .

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

$λ = \frac{h}{m v}$ $lambda = h/(mv)$ ,

where $h = 6.626 \times 10^{- 34} J \cdot s$ $h = 6.626 xx 10^(-34) "J"cdot"s"$ , or $kg \cdot m^{2} /s$ $"kg"cdot"m"^2"/s"$ , is Planck's constant and $m$ $m$ 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:

$λ = \frac{h}{m} \times \sqrt{\frac{π m}{8 k_{B} T}}$ $color(blue)(lambda) = h/m xx sqrt((pim)/(8k_B T))$

$= \sqrt{\frac{π h^{2}}{8 m k_{B} T}}$ $= sqrt((pih^2)/(8mk_B T))$

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

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

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

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

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Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use $u_{a v g} = \sqrt{\frac{8 k_{B} T}{π m}}$ $u_(avg) = sqrt((8k_BT)/(pi m))$

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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))uavg=√8kBTπmu_(avg) = sqrt((8k_BT)/(pi m))

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Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use $u_{a v g} = \sqrt{\frac{8 k_{B} T}{π m}}$ $u_(avg) = sqrt((8k_BT)/(pi m))$