(a) For what K.E. of a neutron, will the associated de Broglie wavelength be `1.40xx 10^(-10)m`?
(b) Also find the de - Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of` (3//2)` kT at 300K

Explanation

(a)` p=h//lambda and K.E = p^(2)//2m = h^(2)/2lambda^(2)m) `
i.e., `K.E. = ((6.626xx10^(-34))/(1.4xx10^(-10)))^(2)xx 1/2 xx 1/(1.67xx 10^(-2))=6.71xx 10^(21)J`
`K.E.= 0.0 419 eV`
(b) `lambda = (h)/sqrt(mnkT), n = 3 ` for a neutron
`therefore lambda = (6.626xx10^(-34))/(sqrt(1.67xx10^(-27)xx3xx1.38xx300xx10^(-23)))`
i.e., `lambda = (6.626xx10^(-34))/(45.54xx10^(-25)) m`
i.e., `lambda = 0.145xx10^(-9) m or lambda = 0.145nm`