For what kinetic energy of a neutron will the associated de-Broglie wavelength be `1.40xx10^(-10)m`?

Explanation

`lambda=1.40xx10^(-10)m`,
`h=6.63xx10^(-34) Js`
`m=1.67xx10^(-27)`kg
(a) de-Broglie wavelength of neutron,
`lambda=(h)/(sqrt(2mK))`
`therefore K=(h^(2))/(2mlambda^(2))`
`=((6.63xx10^(-34))^(2))/(2xx1.67xx10^(-27)xx(1.4xx10^(-10))^(2))`
`therefore K=6.686xx10^(-21)`J
`=(6.686xx10^(-21))/(1.6xx10^(-19))eV`
`therefore K=4.174xx10^(-2)eV`
(b)Average kinetic energy of neutron
Here `k_(B)=1.38xx10^(-23)Jmol^(-1)K^(-1)`
T=300 K
`K=(3)/(2)K_(B)T`
`therefore K=(3)/(2)xx1.38xx10^(-23)xx300`
`therefore K=621xx10^(-23)J`
`therefore` de-Broglie wavelength,
`lambda=(h)/(sqrt(2mK))`
`therefore lambda=(6.63xx10^(-34))/(sqrt(2xx1.67xx10^(-27)xx621xx10^(-23)))`
`therefore lambda=(6.63xx10^(-34))/(45.54xx10^(-25))`
`therefore lambda=0.14558xx10^(-9)m`
`therefore lambda~~0.146` nm