Question
problem 4 please
You know about the "wave/particle duality" of matter. Neutrons are not immune to this base feature of matter? The wavelength of a small particle is one measure of its size. Recall the fundamental relation discovered by de Broglie: lambda = h/p, where h is Plauck's constant and p is the magnitude of the momentum. Let us compare neutron wavelengths against some other sizes and distances of interest. In your calculations, you may use non-relativistic expressions for the neutron momentum and kinetic energy. (a) What is the energy of a neutron whose wavelength equals the "classical diameter" (two times the classical radius) of an electron? (b) What is the energy of a neutron whose reduced wavelength equals the approximate diameter of a uranium-238 nucleus? (c) What is the energy of a neutron whose reduced wavelength approximately equals the distance between nuclei in a typical solid or liquid? This is not a precise question with a precise answer-I just want you to get a handle on these approximate distances and sizes. [Goal: Become familiar with neutron wavelengths, how they change with neutron energy, and how they compare to other small length scales of interest.] Compute the binding energy of the "last neutron, " which is the energy required to remove one neutron, for the following nuclei. (a) ^3H (tritium) (b) ^4He (c) ^9Be (d) ^12C (e) ^13C (f) ^235U (g) ^238U (h) Pick two answers that are significantly different and discuss whether the difference is what you would expect given magic numbers and even/odd considerations. [Goals: Develop understanding of binding energy: calculate it for yourself and thereby increase the chance that you will remember the magnitude of this for a single neutron: see for yourself how "magic numbers" of nucleons lead to more tightly bound nuclei.]
Explanation / Answer
4. a. debroglie's wave equaiton
lambda = h/mv [ m isa mass of particle, lambda is wavelength of the wave, and h is planks constant and v is speed of the particle]
classicval radius of electron, ro = 2.81*10^-15 m
lambda = 2*ro
and if KE of particel is E
E = 0.5mv^2
2E/mv = v
E = 0.5m(h/m*lambda)^2 = 0.5*h^2/m*lambda^2
]for neutron
m = 1.6*10^-27 kg
h = 6.63*10^-34
so, E = 1.739*10^-11 J
b. diameter of U238 nucleus = 2*1.25*10^-15 * (238)^1/3 = 15.492*10^-15 m = lambda
so using
E = 0.5m(h/m*lambda)^2 = 0.5*h^2/m*lambda^2
E = 5.722*10^-13 J
c. Atomicv spacing in average solids, l = 0.4 nm = 0.4*10^-9 m = lambda
using
E = 0.5m(h/m*lambda)^2 = 0.5*h^2/m*lambda^2
E = 8.585*10^-22 J
