The de Broglie relation is

`eq. (1) ->` `P=h/lambda`

Now get the momentum `P` in terms of the kinetic energy `K_E` .

`K_E=1/2 mv^2`

`K_E=P^2/(2m)`

`eq. (2) ->` `P=sqrt(2mK_E)`

Therefore the wavelength can be found by equating `eq. (2)` and `eq. (1)` and solving for `lambda` .

`h/lambda=sqrt(2mK_E)`

`lambda/h=1/sqrt(2mK_E)`

`lambda=h/sqrt(2mK_E)`

Plug in numerical values. You will find the units are much easier if you convert the neutron mass into energy with units of `eV` by` E=mc^2` . Hence,

`lambda=h/sqrt(2mc^2K_E)`

`lambda=(1240 eV*nm)/sqrt(2(940 *10^6 eV)(0.020 eV))=0.20 nm`

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