Answer to Question #123022 in Atomic and Nuclear Physics for Muhammad Adnan Sabir

Question

Neutrons from a nuclear reactor are brought into thermal equilibrium by
repeated collisions in heavy water at T ˆ 300 K. What is the average
energy (in eV) and the typical wavelength of the neutrons? Explain why
they are diffracted when they pass through a crystalline solid.

Answer 1

Answer:

T=300 K

A)What is the average energy and the typical wavelength of the neutrons

We know that,

Average velocity of the neutron by using the Maxwell–Boltzmann distribution of the velocity.

$v=\sqrt{\frac{8KT}{\pi m_{m}}}$

$\therefore$

Energy of the neutron in the thermal equilibrium is is at T=300 K is ,

$E=\frac{1}{2}m_{m}v^{2}=\frac{1}{2}.\frac{8}{\pi }kT=\frac{4}{\pi }kT$

Now,

kT=25.9 mev At T=300 K

$\therefore$

$E=\frac{4}{\pi }kT=\frac{4}{\pi }\times (25.9mev)=32.98\,mev$

So,

$Average\,energy\,(E)=32.98\times 10^{-3}ev$

Now,

We can write energy in terms of wavelength ($\lambda$ ) as,

$\because$ $E=\frac{hc}{\lambda }$

$\implies$ $\lambda =\frac{hc}{E}=\frac{1240ev.nm}{32.98\times 10^{-3}ev}=37.60\times 10^{3}\,nm=37.60\,\mu m$

B) Explain why they are diffracted when they pass through a crystalline solid.

Now,

The de-Broglie wavelength ($\lambda$ ) of the neutron with velocity (v )is,

$\lambda =\frac{h}{m_{m}v}\\$

Where, $v=\sqrt{\frac{3K_{B}T}{\pi m}}$

Here, maximum of the distribution of neutrons occurs.

Now,

The De -Broglie wavelength of the neutrons at T=300 k is of the same other , i.e similar to the atomic spacing , thus permitting the diffraction measurement,when we pass these through the crystalline solid.

the absence scattering of neutron makes it more special because in this nuclear scattering of neutron results from the intersection with the nucleons rather than electrons , so these us the information about magnetic moments not about atomic number.