A neutron in a reactor has kinetic energy of approximately 0.020 eV. Calculate the wavelength of this neutron.

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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`


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

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




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

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