Team Quanta presents all possible short questions of BS Physics book Solid State Physics – I’s Chapter#2: Crystal Binding and Elastic Constants for the students.
Q.1 Why X-rays are used for crystal structure analysis?
Answer. X- rays are used for producing diffraction effect in crystal due to the fact that wavelength of X-rays is comparable to inter atomic distance in actual crystal, for example, X-rays produced by electrons through a potential V have a wavelength given by,
eV=hv
![\[\lambda=\frac{hc}{eV}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-12317796debd793da33c56671186b5e1_l3.png "Rendered by QuickLaTeX.com")
For , the wavelength is comparable to inter atomic spacing of an actual crystal. On the other hand radiation of wavelength shorter than that of X-rays, are diffracted through very small angle which cannot be conveniently measured. It is only X-rays which are scattered elastically without change of wavelength by charged particles of atoms.
Q.2 Why visible light cannot be used to study diffraction from crystal?
OR
Why visible light cannot be used to study crystal structure?
Answer. According to Bragg law,
![\[2d\sin{\theta=n\lambda}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-07fe1762ce3ad81159f5e2aab2dd8793_l3.png "Rendered by QuickLaTeX.com")
The maximum wavelength that can be used for study of crystal structure is 2d. Since this spacing in crystal is of order of whereas wavelength of visible light is which is very large as compared to 2d. Therefore visible light cannot be used for study of crystal structure.
Q.3 Which one is best method out of three crystal diffraction methods and why?
Answer. The powder method can be applied to any type of matter with crystalline arrangement as it does not require single crystal. Therefore powder method is most widely used in investigation of metals and alloys and in applied work.
Q.4 Prove that in determining lattice parameters, the greater the diffractions angle, the greater is accuracy.
Answer. Differentiating Bragg law 
we find that
![\[\frac{\Delta d}{d}\propto cos\theta \Delta \theta\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-9559110eae524480d8d0b47bdcbad9f0_l3.png "Rendered by QuickLaTeX.com")
This shows that for greater angle, 
is smaller and hence greater is accuracy in measuring d.
Q.5 What do you mean by reciprocal lattice?
Answer. Reciprocal lattice is a regular and periodic arrangement of points which represents the slope and inter planar spacing of different sets of parallel planes of direct lattice. A reciprocal lattice is spanned b translational vector 
satisfying the conditions:
![\[\vec{a}.\ \vec{G}=2\pi h,\ \ \ \ \ \vec{b}.\ \vec{G}=2\pi k,\ \ \ \ \ \vec{c}.\vec{G}=2\pi l\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-6b0b96ff9b70183113a74f2187ea47c1_l3.png "Rendered by QuickLaTeX.com")
Where are primitive translation vectors of direct lattice.
Q.6 How is reciprocal lattice vector constructed?
Answer. A reciprocal lattice vector can be constructed in following manner:
- Take a point in direct lattice as origin.
- From this point draw normal to every set of planes in direct lattice.
- Equal the length of each normal to reciprocal of inter planer spacing for its particular set of planes.
- Place points at end of each normal.
The set of regular and periodic points so obtained give a reciprocal lattice.
Q.7 What is importance of diffraction phenomena of light?
Answer. Diffraction of light is used;
- To find wavelength of any monochromatic light.
- To discover crystal structure of different compounds.
- To know ionic spacing in different crystal.
Q.8 Why neutral crystals are used for X-ray diffraction instead of diffraction grating?
Answer. X-rays are electromagnetic waves of very short wavelength of order of This wavelength is too small to draw an ordinary grating plate and does not allow X-rays to diffract through it. Bragg discovered that natural crystal have spacing comparable to wavelength of X-rays. This spacing is sufficient for diffraction of X-rays through natural crystals.
Q.9 Define atomic scattering factor.
Answer. Atomic scattering factor is defined as ratio of amplitude of radiation scattered from atom to amplitude of radiation scattered from an electron.
![\[f=\frac{amplitude\ of\ radiation\ scattered\ from\ atom}{amplitude\ of\ radiation\ scattered\ from\ electron}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-3153d7c55e9e1d657159a44cd95113e5_l3.png "Rendered by QuickLaTeX.com")
Q.10 State Laue diffraction conditions by explaining all the terms used.
Answer. Laue diffraction conditions are;
![\[2a\sin{\theta\cos{\alpha=h\lambda}}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-5ea340b73f2fd29339e918cefb432132_l3.png "Rendered by QuickLaTeX.com")
![\[2b\sin{\theta\cos{\beta=k\lambda}}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-e30bb43a39e1b8be3a27361c1fe2b8bf_l3.png "Rendered by QuickLaTeX.com")
![\[2c\sin{\theta\cos{\gamma=l\lambda}}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-4757d3195064c56e3f597928366e8765_l3.png "Rendered by QuickLaTeX.com")
Here is angle which the incident beam makes with reflecting plane and λ is wavelength of incident radiation. are angles between normal to reflecting plane and are axes respectively.
Q.11 How Bragg law can be deduced from Laue diffraction condition?
Answer. From Laue equations,
![\[\cos{\alpha=\frac{\lambda}{2\sin{\theta}}}\ \frac{h}{a},\ \ \ \cos{\beta=\frac{\lambda}{2\sin{\theta}}\ }\frac{k}{b},\ \ \ \cos{\gamma=\frac{\lambda}{2\sin{\theta}}}\ \frac{l}{c}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-04fdcc85fe41c1a2b292f89cee2a1f2e_l3.png "Rendered by QuickLaTeX.com")
Example: Direction cosines are 
respectively. If d is spacing between two adjacent planes of set , then
![\[d=\frac{a}{h}\cos{\alpha}=\frac{b}{k}\cos{\beta}=\frac{c}{l}\cos{\gamma}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-5030595a344aaa540c242db7fc41f1b5_l3.png "Rendered by QuickLaTeX.com")
According to Laue equations,
![\[\frac{a}{h}\cos{\alpha}=\frac{b}{k}\cos{\beta}=\frac{c}{l}\cos{\gamma}=\frac{\lambda}{2\sin{\theta}}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-14f815b5e0941fe744890b029b77a470_l3.png "Rendered by QuickLaTeX.com")
So Laue equations can be put inform,
![\[d=\frac{\lambda}{2\sin{\theta}}\]](https://quantapublisher.com/wp-content/ql-cache/quicklatex.com-79d34e6300076c37abb868b54aebb623_l3.png "Rendered by QuickLaTeX.com")
Introducing common factor n, we have
This is Bragg equation.
Q.12 Which method is best for checking orientation of crystals in solid stae experiments?
Answer. In Laue method, diffraction pattern consists of series of spots showing symmetry of crystals. If a crystal with four fold axis of symmetry with axis parallel to beam, Laue pattern will show 4-fold symmetry. This feature makes Laue method especially useful for checking the orientation of crystal in solid state experiments.
Q.13 Which method is best for determination of size of unit cell?
Answer. The rotating crystal method is best suited method for structure determination when single crystal specimen is available. If rotation photographs are taken separately about three axes translation vectors and hence dimensions of unit cell of crystal is determined. Thus rotating crystal method is a very powerful tool for determining size of unit cell.
Review Questions
