A. For what kinetic energy of neutron will the associated de Broglie wavelength be `1.40xx10^(-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 `300 K`. Given the mass of neutron`=1:66xx10^(-27)kg` and `k=1.38xx10^(-23)Jkg^(-1)`.`

Explanation

(a) Given , de Broglie wavelength , `lamda=1.40 xx10^(-10)m`
Kinetic energy of neutron is given by `E_k=h^2/(2lamda^2m)`
`=(6.62xx6.62xx10^(-68))/(2xx1.4xx1.4xx10^(-20)xx1.67xx10^(-27))`
`" "[m=1.675xx10^(-27) kg]`
`=6.69xx10^(-21)J`
or `K_k=(6.69xx10^(-21))/(1.6xx10^(-19))eV`
`=4.18xx10^(-2)eV`
(b) The de Broglie wavelength of neutron is given by
`lamda=h/(sqrt2mkT)`
Here, temperature , T = 300 K
k = Boltzmann constant `=1.38xx10^(-23)m^2"kg s"^(-2)K^(-1)`
`implieslamda=(6.62xx10^(-34))/(sqrt(3xx1.67xx10^(-27)xx1.38xx10^(-23)xx3000))`
`=(6.62)/(45.5)xx10^(-9)`
`=0.145xx10^(-9) = 0.145nm`