Question

17. What is the de Broglie wavelength of a neutron that has fallen $1.0\mathrm{m}$in a vacuum chamber, starting from rest?

Short answer

The de Broglie wavelength of the neutron that has fallen $1.0m$in a vacuum chamber is$\lambda =90\mathrm{nm}$

Step-by-step solution

Given information 

the de Broglie wavelength of a neutron that has fallen $1.0m$in a vacuum chamber,

calculation 

The neutron has traveled through the gravitational potential energy difference of

$E=mg\Delta y$

where $\Delta y=1.0\mathrm{m}$. This potential energy difference is converted into neutron's kinetic energy, so

$T=mg\Delta y$

We also know that the kinetic energy is related to momentum through

$T=\frac{p^2}{2m}$

Therefore

$\frac{p^2}{2m}=mg\Delta y$

yielding

$p=m\sqrt{2g\Delta y}.$

De Broglie wavelength is then simply given by

$+\lambda -\frac{h}{p}=\frac{h}{m\sqrt{2g\Delta y}}.$

Remember that $h=6.626·{10}^{-34}\mathrm{Js}$, and that the mass of the neutron is $m=1.67·{10}^{-27}\mathrm{kg}$. With these numerical values we get

$\lambda =90\mathrm{nm}$