The de-Brogile wavelength of a neutron at `927^(@)`C is `lamda`. What will be its wavelength at `27^(@)`C?

Options

A. `lamda`
B. `lamda/2`
C. `(3lamda)/(2)`
D. `2lamda`

Answer

D

Explanation

The K.E. `= 3/2 KT`
`therefore 1/2mv^2 = 3/2 KT therefore m^2v^2 = 3KTm`
`therefore mv = sqrt(3mKT)`
`therefore` The de broglie wavelength `(lamda)` is given by
`therefore lamda = (h)/(mv) = (h)/(sqrt(3mKT)) " or " lamda prop 1/sqrtT`
`therefore lamda_2/lamda_1 = sqrt(T_1/T_2) = sqrt((927+273)/(27+273)) = sqrt(1200/300) = 2`
`therefore lamda_2 = 2lamda_1 " or " lamda_2 = 2lamda`