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(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de$-$Broglie wavelength is given by $\lambda = \frac{h}{m

Vedclass · Accepted Answer

$0.24$

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(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de$-$Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Substituting $v_{rms}$ for $v$, we get $\lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3kTm}}$.Given values: $h = 6.63 	imes 10^{-34} \ J \cdot s$, $k = 1.38 	imes 10^{-23} \ J/K$, $T = 400 \ K$, and $m = 4.64 	imes 10^{-26} \ kg$.Calculating the denominator: $\sqrt{3 	imes 1.38 	imes 10^{-23} 	imes 400 	imes 4.64 	imes 10^{-26}} = \sqrt{7.68768 	imes 10^{-45}} \approx 2.77 	imes 10^{-22} \ kg \cdot m/s$.Now, $\lambda = \frac{6.63 	imes 10^{-34}}{2.77 	imes 10^{-22}} \approx 2.39 	imes 10^{-11} \ m$.Converting to $\mathring{A}$s: $2.39 	imes 10^{-11} \ m = 0.239 \ \mathring{A} \approx 0.24 \ \mathring{A}$.

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