What is the de-Broglie wavelength of nitrogen molecule in air at 300K ? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen `=14.0076U`) Plank's constant=`6.63xx10^(-34)Js` , Boltzmann constant `=1.38xx10^(-23)JK^(-1)`

Explanation

Let v = rms speed of `N_2` molecule at 300 K.
or ` v = sqrt((3kT)/(m) )`
or ` v = sqrt( (3 xx 1.38 xx 10^(-23) xx 300 )/(28.0152 xx 1.67 xx 10^(-27 ) ) ) `
` = 5.15 xx 10^2 ms^(-1)`
Using the formula `lamda = (hc)/(mv) ` , we get
`lamda = (h)/(mv)`
` = (6.62 xx 10^(-34) )/(28.0152 xx 1.67 xx 10^(-27) xx 5.15 xx 10^2 ) `
` = 2.57 xx 10^(-11) m`