The average kinetic energy of a neutron at 300 K is `(3)/(2)`kT. Find its de Broglie wavelength.

Explanation

de Broglie wavelength
`lamda = (h)/(sqrt(2mE)) = (h)/(sqrt(3mkT)) [ :' E = (3)/(2) kT]`
`= (6.626 xx 10^(-34))/((3 xx 1.67 xx 10^(-27) xx 1.38 xx 10^(-23) xx 300)^(1//2))`
`[ :' k = 1.38 xx 10^(-23)]`
`= 1.456 Å`