The de Broglie wavelength of a molecules of thermal energy `KT` ( `K` is Boltzmann constant and `T` is absolute temperature) is given by

Options

A. `lamda = sqrt((h)/(2mK_BT))`
B. `lamda=(h)/(4m^2K_B^2 T^2)`
C. `lamda=hsqrt(2mK_BT)`
D. `lamda = (h)/(sqrt(2mK_BT))`

Answer

D

Explanation

For charged particles , `lamda = (h)/(sqrt(2meV))`
But for unchanged particles like neutrons, atoms molecules, etc.
`K.E. = 1/2mv^2 = K_B T therefore v = sqrt((2K_BT)/(m))`
de broglie wavelength
`lamda = (h)/(mv) = (h)/(sqrt((m^2xx2K_BT)/(m)))`
`therefore lamda = (h)/(sqrt(2K_BmT))`