Calculate the de Broglie wavelength of a CO_2 | vedclass

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Question

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.First,calculate the mass of one $CO_2$ molecule in $kg$:$Molar \, mass

Vedclass · Accepted Answer

$2.063 	imes 10^{-11} \, m$

Vedclass · Answer

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.First,calculate the mass of one $CO_2$ molecule in $kg$:$Molar \, mass = 44 \, g/mol = 0.044 \, kg/mol$.$Mass \, of \, one \, molecule \, (m) = \frac{0.044 \, kg/mol}{6.022 	imes 10^{23} \, mol^{-1}} \approx 7.306 	imes 10^{-26} \, kg$.Now,substitute the values into the de Broglie equation:$\lambda = \frac{6.626 	imes 10^{-34} \, J \cdot s}{(7.306 	imes 10^{-26} \, kg) 	imes (440 \, m/s)}$.$\lambda = \frac{6.626 	imes 10^{-34}}{3.2146 	imes 10^{-23}} \, m$.$\lambda \approx 2.061 	imes 10^{-11} \, m$.Rounding to the nearest option,the correct answer is $2.063 	imes 10^{-11} \, m$.

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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