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Question: (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exer...

(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn​=1.675×1027^{27} kg)

(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27∘c^{\circ }c ). Hence, explain why a fast neutron beam needs to be thermalized with the environment before it can be used for neutron diffraction experiments.

(c) Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1A∘1A^\circ which is of the order of interatomic spacing in the lattice) (mem_e ​= 9.11×10−31^{-31} kg)

Explanation

Solution

In the question Kinetic energy and Mass of neutron is given. Now, we have to represent kinetic energy in a form by which we can find the wavelength of the body. After representing, we will calculate the wavelength for the neutron and compare whether it is suitable for a diffraction experiment or not. For the second question, we should convert the Celsius scale to Fahrenheit scale and then calculate the average kinetic energy and with that we have to find the value of wavelength again and compare it.

Complete answer:
Answer for question (A),
Here in the first question Kinetic energy of the neutron is given as,
E=150eV=150×1.6×10−19JE=150eV=150\times 1.6\times {{10}^{-19}}J
Mass of the neutron is also given as,
m=1.675×10−27kgm=1.675\times {{10}^{-27}}kg
We know that kinetic energy of neutron can be represented as,

E=12mv2  or  mv=2EmE=\dfrac{1}{2}m{{v}^{2}}\ \,\text{or}\ \,mv=\sqrt{2Em}
Therefore, the wavelength of the beam of neutron would be,
∴λ=hmv=h2Em=6.63×10−342×150×1.6×10−19×1.675×10−27=2.33×10−12m\therefore \lambda =\dfrac{h}{mv}=\dfrac{h}{\sqrt{2Em}}=\dfrac{6.63\times {{10}^{-34}}}{\sqrt{2\times 150\times 1.6\times {{10}^{-19}}\times 1.675\times {{10}^{-27}}}}=2.33\times {{10}^{-12}}m
Therefore, a neutron beam of 150eV is not suitable for diffraction experiment as it does not have the required wavelength, the wavelength required must be 100 times greater than the wavelength of a beam of neutron that is ∼1A∘(=10−10m)\sim1{A^\circ}(={{10}^{-10}}m).

Answer for question (B)
Here the temperature given is in Celsius scale but we require it in Fahrenheit scale so,
T=27+273=300KT=27+273=300K
We know that Boltzmann constant is,
k=1.38×10−23Jmol−1K−1k=1.38\times {{10}^{-23}}Jmo{{l}^{-1}}{{K}^{-1}}
We know that at an absolute temperature T the average kinetic energy of neutron is given as,
E=32kTE=\dfrac{3}{2}kT, Here k is the Boltzmann constant.
Therefore, the wavelength of the beam of neutron is,
λ=h2Em=h3mkT=6.63×10−343×1.675×10−27×1.38×10−23×300=1.45×10−10m\lambda =\dfrac{h}{\sqrt{2Em}}=\dfrac{h}{\sqrt{3mkT}}=\dfrac{6.63\times {{10}^{-34}}}{\sqrt{3\times 1.675\times {{10}^{-27}}\times 1.38\times {{10}^{-23}}\times 300}}=1.45\times {{10}^{-10}}m
A high energy of a beam of neutron must be neutralized before using it for diffraction, because the wavelength of neutron is comparable to inter atomic spacing (∼1A∘)(\sim1{A^\circ}).

Answer for question (C)
For the same wavelength an x-ray probe has a greater wavelength than an electron probe.
We know that we can get the kinetic energy for an electron is,
E=12mv2  or  mv=2Em  and  λ=hmv=h2EmE=\dfrac{1}{2}m{{v}^{2}}\ \,\text{or}\ \,mv=\sqrt{2Em}\ \,\text{and}\ \,\lambda =\dfrac{h}{mv}=\dfrac{h}{\sqrt{2Em}}

∴E=h22λ2m=(6.63×10−34)22×10−20×9.11×10−31J=(6.63×10−34)22×9.11×10−51×1.6×10−19eV=150.6eV\therefore E=\dfrac{{{h}^{2}}}{2{{\lambda }^{2}}m}=\dfrac{{{(6.63\times {{10}^{-34}})}^{2}}}{2\times {{10}^{-20}}\times 9.11\times {{10}^{-31}}}J=\dfrac{{{(6.63\times {{10}^{-34}})}^{2}}}{2\times 9.11\times {{10}^{-51}}\times 1.6\times {{10}^{-19}}}eV=150.6eV
And for photons wavelength is,
λ=1A∘=10−10m\lambda =1{A^\circ}={{10}^{-10}}m
Therefore, the energy of photons is,

E=hv=hcλ=6.63×10−34×3×10810−10×1.6×10−19eV=12.4×103eVE=hv=\dfrac{hc}{\lambda }=\dfrac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{{{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV=12.4\times {{10}^{3}}eV
Hence it is proved that for the same wavelength x ray probe has a greater energy than an electron probe.

Note:
Remember the formula for Kinetic energy and average kinetic energy and the energy of photons. Remember the formula how average kinetic energy is represented. While doing the problems check in which parameter temperature is given because we have to find each and every result in Fahrenheit scale.