A neutron with kinetic energy `T=25 eV` strikes a stationary deuteron (heavy hydrogen nucleus). Find the de Broglie wavelength of both particles in the frame of their centre of intertia.

Explanation

We shall use `M_(0)~~2M_(n)`. The `CM` is moving with velocity
`V=(sqrt(2M_(n)T))/(3M_(n))=sqrt((2T)/(9M_(n)))`
with respect to the Lab frame. In the `CM` frame the velocity of neutron is
`V'_(n)=V_(n)-V=sqrt((2T)/(M_(n)))-sqrt((2T)/(9M_(n)))=sqrt((2T)/(M_(n))).(2)/(3)` and `lambda'_(n)=(2piħ)/(M_(n)V'_(n))=(3piħ)/(sqrt(2M_(n)T))`
Substitution gives `lambda'_(n)= 8.6p m`
Since the momenta are equal in the `CM` frame the Broglie wavelength will also be equal. If we do not assume `M_(d)~~2M_(n)` we shall get
`lambda_(n)=(2 piħ(1+M_(n)//M_(d)))/(sqrt2M_(n)T)`