Find the de Broglie wavelength associated wit | vedclass

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(B) The de Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{3mk_BT}}$.Here,$m$ is the mass of a Helium atom.$m = \frac{4 	imes 1

Vedclass · Accepted Answer

$7.3 	imes 10^{-11} \ m$

Vedclass · Answer

(B) The de Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{3mk_BT}}$.Here,$m$ is the mass of a Helium atom.$m = \frac{4 	imes 10^{-3} \ kg}{6.023 	imes 10^{23}} \approx 0.664 	imes 10^{-26} \ kg$.The temperature $T = 27^{\circ}C = 27 + 273 = 300 \ K$.Substituting the values into the formula:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes (0.664 	imes 10^{-26}) 	imes (1.38 	imes 10^{-23}) 	imes 300}} \ m$.Calculating the denominator:$\sqrt{3 	imes 0.664 	imes 1.38 	imes 300 	imes 10^{-49}} = \sqrt{824.112 	imes 10^{-49}} \approx \sqrt{82.41 	imes 10^{-48}} \approx 9.08 	imes 10^{-24}$.$\lambda \approx \frac{6.63 	imes 10^{-34}}{9.08 	imes 10^{-24}} \approx 0.73 	imes 10^{-10} \ m = 7.3 	imes 10^{-11} \ m$.

Find the de Broglie wavelength associated with a Helium $(He)$ atom in Helium gas at room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure.

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