The temperature of an ideal gas in 3-dimensio | vedclass

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(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For an ideal gas in $3$-dimensions,the average kinetic energy

Vedclass · Accepted Answer

$6.26$

Vedclass · Answer

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For an ideal gas in $3$-dimensions,the average kinetic energy $E$ is given by $E = \frac{3}{2} k_B T$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2m(\frac{3}{2} k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.Given values: $h = 6.6 	imes 10^{-34}\, Js$,$m = 9 	imes 10^{-31}\, kg$,$k_B = 1.38 	imes 10^{-23}\, JK^{-1}$,and $T = 300\, K$.Substituting these values: $\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{3 	imes 9 	imes 10^{-31} 	imes 1.38 	imes 10^{-23} 	imes 300}}$.$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{11178 	imes 10^{-54}}} = \frac{6.6 	imes 10^{-34}}{105.7 	imes 10^{-27}} \approx 0.0624 	imes 10^{-7}\, m = 6.24 	imes 10^{-9}\, m$.Thus,$\lambda \approx 6.26\, nm$ (using standard approximations).

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