The de Broglie wavelength of an neutron corresponding to root mean square speed at `927^(@)C` is `lambda`. What will be the de Broglie wavelength of the neutron corresponding to root mean square speed at `27^(@)C`?

Options

A. `(lambda)/(2)`
B. `lambda`
C. `2 lambda`
D. `4 lambda`

Answer

C

Explanation

`K.E.` of neutron `E=(3)/(2)KT`
`lambda_(d)=(h)/(p)=(h)/(sqrt(2mE))=(h)/(sqrt(2mxx(3)/(2)kT))`
`rArr lambda_(2)=lambdasqrt(((927+273))/(27+273))= 2 lambda`