Calculate the lambda of CO_2 molecule moving | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $440 \ m/s.$

Similar Questions

If the kinetic energy of an electron is $1.25 \times 10^{-28} \ J$,then what will be the de-Broglie wavelength of the electron? (Mass of electron $= 9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ Js$)

Which of the following represents the de Broglie equation?

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $2.0 \times 10^{-25} \ J$,the wavelength of the electron in $nm$ is approximately (in $.3$)

The de-Broglie wavelength of a particle with mass $1 \, g$ and velocity $100 \, m/sec$ is

What is the de-Broglie wavelength of an electron,in a hydrogen atom,moving in an orbit having a maximum magnetic quantum number $m = +2$ in units of $\mathring{A}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module