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

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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