For what kinetic energy of a neutron, will the associated de-Broglie wavelength be `1.32xx10^(-10)m` ? Given that mass of a neutron`=1.675xx10^(-27)kg`.

Explanation

It is given that `lamda=1.32xx10^(-10)m and m_(n)=1.675xx10^(-27)kg`
`because lamda=(h)/(sqrt(2m_(n)K)) implies K=(h^(2))/(2m_(n)lamda^(2))`.
`therefore`Kinetic energy of neutron `K=((6.63xx10^(-34))^(2))/(2xx(1.675xx10^(-27))xx(1.32xx10^(-10))^(2))=7.53xx10^(-21)`.