Calculate de Broglie wavelength for an average helium atom in a furnace at 400 K. Given mass of helium`=4.002`amu.

Explanation

De Broglie wavelength, `lamda=(h)/(p)`
Mean KE of helium atom `=(3kT)/(2)`
or `E=(3xx(1.38xx10^(-23))xx400)/(2)`
In terms of momentum, the energy is given by
`E=(p^2)/(2m)` or `p^2=2mE` or `p=sqrt(2mE)`
`p=sqrt(2xx(4.002xx1.66xx10^(-27))xx3xx(1.38xx10^(-23))xx200)` (where mass of helium`=4.002 a m u =4.002xx1.66xx10^(-27)kg`)
Now, `lamda=(6.625xx10^(-34))/(p)`
Substituting the value of p and solving it, we get `lamda=0.63xx10^(-10)`m