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)De Broglie wavelength of the neutron , `lambda=1.40xx10^(-10) m`
Mass of a neutron , `m_n=1.66xx10^(-27) kg`
Planck's constant, `h=6.6xx10^(-34)`Js
Kinetic energy (K) and velocity (v) are related as : `K=1/2 m_n v^2` …(1)
De Broglie wavelength `(lambda)` and velocity `(v)` are related as : `lambda=h/(m_n v)`....(2)
Using equation (2) in equation (1), we get
`K=1/2 (m_nh^2)/(lambda^2 m_n^2)=h^2/(2lambda^2m_n)`
`=((6.63xx10^(-34))^2)/(2xx(1.40xx10^(-10))^2 xx 1.66xx 10^(-27))=6.75xx10^(-21) J`
`=(6.75xx10^(-21))/(1.6xx10^(-19))=4.219xx10^(-2) eV`
Hence, the kinetic energy of the neutron is `6.75xx10^(-21) J or 4.219xx10^(-2)` eV
(b)Temperature of the neutron, T=300 K
Boltzmann constant, k=`1.38xx10^(-23) kg m^2 s^(-2) K^(-1)`
Average kinetic energy of the neutron :
`K'=3/2 kT`
`=3/2xx1.38xx10^(-23)xx300=6.21xx10^(-21) J`
The relation for the de Broglie wavelength is given as :
`lambda'=h/sqrt(2K'm_n)`
Where, `m_n=1.66xx10^(-27)` kg
`h=6.6xx10^(-34) Js`
`K'=6.75 xx10^(-21) J`
`therefore lambda'=(6.63xx10^(-34))/(sqrt(2xx6.21xx10^(-21)xx1.66xx10^(-27)))=1.46xx10^(-10) m =0.146 nm`
Therefore, the de Broglie wavelength of the neutron is 0.146 nm