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. Here `lamda=1.40lamdaxx10^(-10)m`,`m=1.66xx10^(-27)kg`
The energy of neutron is given by
`E=(h^2)/(2mlamda^2)`
`=((6.62xx10^(-34))^2)/(2xx1.66xx10^(-27)xx(1.40xx10^(-10))^2)`
`=6.735xx10^(-21)J`
B. Here `k=1.38xx10^(-23)Jkg^(-1)` and `T=300K`
Therefore, energy of neutron
`E=(3)/(2)xx=(3)/(2)xx1.38xx10^(-23)xx300=6.21xx10^(-21)J`
Now, `lamda=(h)/(sqrt(2mE))=(6.62xx10^(-34))/(sqrt(2xx1.66xx10^(-27)xx6.21xx10^(-21)))`
`=1.46xx10^(-10)=1.46A`