What is the kinetic energy (in eV) of a Broglie wavelength of 1.4x10^-10m. Knowing the mass of a neutron is 1.67x10^-27 kg.

Question

michele_laino · Answer

please, note that the kinetic energy E  of a not-relativistic neutron whose mass is M, is equals to:
$$E=\frac{ p ^{2} }{ 2M }$$
where p is the momentum of our neutron. By De Broglie relationship, we can write:
$$\lambda=\frac{ h }{ p }$$
where lambda is the wave length of our neutron, and h is the Planck constant, namely h=6.62*10^-27 erg*sec

michele_laino · Answer

so after substitution into relationship for E, we get:
$$E=\frac{ h ^{2} }{ 2M \lambda ^{2} }$$

michele_laino · Answer

now, we have to insert your numerical data into equation above, in order to get the kinetic energy E, so we have:

michele_laino · Answer

$$E \approx 6*10^{-14} erg$$

michele_laino · Answer

or $$E \approx 6*10^{-21} Joule$$
now, keep in mind that 
$$1eV=1.6*10^{-19 }Joule$$
we get:
$$E \approx 3.2*10^{-2}eV$$