Calculate the lambda of CO_2 molecule moving | vedclass

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(D) The wavelength is calculated using the de-Broglie equation: $\lambda = \frac{h}{m 	imes u}$.First,calculate the mass of one $CO_2$ molecule: $m =

Vedclass · Accepted Answer

$\lambda = 2.06 	imes 10^{-11} \ m$

Vedclass · Answer

(D) The wavelength is calculated using the de-Broglie equation: $\lambda = \frac{h}{m 	imes u}$.First,calculate the mass of one $CO_2$ molecule: $m = \frac{44 \ g/mol}{6.022 	imes 10^{23} \ mol^{-1}} = 7.306 	imes 10^{-23} \ g = 7.306 	imes 10^{-26} \ kg$.Now,substitute the values into the equation: $\lambda = \frac{6.626 	imes 10^{-34} \ J \cdot s}{(7.306 	imes 10^{-26} \ kg) 	imes (440 \ m/s)} = 2.06 	imes 10^{-11} \ m$.Thus,the wavelength of the $CO_2$ molecule is $2.06 	imes 10^{-11} \ m$.

Calculate the $\lambda$ of $CO_2$ molecule moving with a velocity $440 \ m/s.$

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Calculate the de Broglie wavelength of a $CO_2$ molecule moving with a velocity of $440 \, m/s$. (Molar mass of $CO_2 = 44 \, g/mol$)

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