## de Broglie wavelength

$\lambda=\frac{h}{p}$

Juliana Logan 1A
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

### de Broglie wavelength

Question 4 from the fall 2013 midterm:
During spontaneous fission uranium-235 emits a neutron with a mass of approximately 1.00866 g/mol. If one of these neutrons has an energy of 4.84 MeV, what is its de Broglie wavelength?
I'm confused on what the 1.00866 g/mol is used for, is it to find the mass of an neutron, but I thought that was usually given? It isn't really explained in the solutions, it just goes from 1.00866 g/mol to the mass of a neutron, so could someone please explain this?

Gloria Quach 1H
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

### Re: de Broglie wavelength

I believe that the mass of a neutron usually isn't provided on the formula sheet; the mass of an electron would be typically provided on the sheet.
The given 1.00866 g/mol was then converted into the mass of a single neutron atom. In other words, 1.00866 g/mol was divided by 10^(-3) to convert into kg/mol and then divided by Avogadro's number (6.02214 x 10^23) to get the mass of a single neutron atom in kilograms.

304655635
Posts: 9
Joined: Fri Sep 25, 2015 3:00 am

### Re: de Broglie wavelength

What led de-Broglie to think that the material particles may also show wave nature?