# Calculate Wavelength of 200Kev Electron

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```					MatE. 297                                                        Instructor: Zhen Guo

Quiz #5 (04/13)

Nano Material Characterization

1. Calculate the X-ray angle for Si (111) and (200) peak, giving the lattice
parameter of Si is 5.43A and X-ray wavelength is 1.54A (Cu Ka)

Bragg's Law: 2dsin=nSin-1 (2d)

For Si a=5.43A,
a       5.43                                                   1.54
d 111                   3.125 A , 111  Sin 1 (        )  Sin 1 (           )  14 .2 
h k l
2    2 2
3                              2d111               2 * 3.125
a       5.43                                                    1.54
d 200                   2.715 A ,  200  Sin 1 (        )  Sin 1 (           )  16 .5
h k l
2    2 2     2                               2d 200              2 * 2.715

2. Calculate the wave length and velocity for electron beam in TEM with
acceleration energy of 200KV and 1.5MV

De Broglie Relationship: =h/P, P=(2mE)0.5, v=P/m

For 200kV, E=200keV=200*103*1.6*10-19=3.2*10-14J
h            6.63  1034
                                        2.75  1012 m  0.0275A
30         14
2mE    2  0.91  10  3.2  10
2me 2.41  10 22
v                  30
 2.65  10 8 m / s 0.88 times of speed of light
m    0.91  10

For 1.5MV, E=1.5MeV=1.5*106*1.6*10-19=2.4*10-13J
h            6.63  1034
        
30         13
 1.00  1012 m  0.01A
2mE    2  0.91  10  2.4  10
2me 6.61  10 22
v                  30
 7.26  10 8 m / s 2.42 times of speed of light!!!
m    0.91  10

This number is awfully large. We all know that nothing can exceed the speed of
light. So what's wrong? This is because when electron speed is very high
approaching speed of light, we have to take The Theory of Relativities into
account. According to Relativities Quantum Mechanics, the electron mass is no
longer a constant, but rather a function of energy. Also the way to calculate the
velocity is also different. We will not discuss details here as it is beyond the
scope of this class but this is just a reminder to always check your results and
see whether it makes sense or not.

See the table below from David Williams and C. Barry Carter's book
"Transmission Electron Microscopy"

Accelerating    Nonrelativistic     Relativistic          Mass        Velocity
8
Voltage (KV)    Wavelength (nm)     Wavelength (nm)       (X me0)     (X10 m/s)
100             0.00386                0.0037       1.196         1.644
120             0.00352              0.00335        1.235         1.759
200             0.00273              0.00251        1.391         2.086
300             0.00223              0.00197        1.587           2.33
400             0.00193              0.00164        1.783         2.484
1000             0.00122              0.00087        2.957         2.823

3. Using the concept of reciprocal lattice to explain how to determine grain size
from X-ray diffraction peak width.

According to Bragg's law, there is only 1 theta angle corresponding to one plan.
So for an infinite large crystalline, the diffraction pattern should be a single line.
In reality, every peak has its width due to multiple grain size. In reciprocal space,
lattice points are opposite with its physical shape. For example, a thin film will
lead to a reciprocal lattice point with stick shape while nano fiber will cause a 2-D
planar reciprocal lattice point. As a consequence, smaller grains will leads to
larger reciprocal lattice point and therefore wider diffraction peak. Hence, width
or half width of peak provide information of grain size and we can extract or
back-calculate grain size from it.

Grain Size
Decrease

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