Energy of neutrons in crystal diffraction

The spacing of the atoms of a crystal is 159 pm. A monoenergetic beam of neutrons directed normally at the surface of the crystal undergoes first order diffraction at an angle of 58° from the normal. What is the energy of each of the neutrons?

For this question, I used the Bragg condition as if the neutron beam were an X-ray beam. Since 58º was measured from the normal, the angle I should substitute into the formula would be $90º-58º=32º$. $$2dsin{\phi}=m\lambda$$ $$2(159 \times 10^{-12} m)(sin{32º})=(1)\lambda$$ Solving for $\lambda$ gave me $1.69 \times 10^{-10}$ m. Then, $$E=hf=h\frac{c}{\lambda}=(4.14 \times 10^{-15} eV•s)\frac{3 \times 10^8 m/s}{1.69 \times 10^{-10} m}$$ which gave me 7370 eV.

The answer to the question is 0.045 eV. Where did I go wrong?

You used the energy-wavelength relation for photons

$$E=h\nu$$

which is valid for neutrons as well, but with some implicit conversions.

Here, since neutrons don't move with speed c, you can't use that. Instead, you use the velocity implicit in the de Broglie equation.

In the future, try this:

For neutrons, use the energy approximation like this:

$$\lambda (in\ angstrom)=\frac {0.286}{\sqrt {E ( in\ eV)}}$$