使用Scherrer公式计算晶粒尺寸

Question

我试图用Scherrer公式计算晶粒尺寸，但我一直在半高宽。

import numpy as np 
#import math 

k = 0.94 
wave_length = 1.5406e-10 

data = np.genfromtxt("G3.txt") 

indice = np.argmax(data[:,1]) 
peak = (data[indice, :]) 
#D = (k*wave_length)/(beta*cos((math.radian(theta)) 

信息：Scherrer equation，Full width at half maximum，Related question

Answer 1

我不知道如何在Python解决这个问题（至少此时）。所以我用Matlab做了。

clear all 
    clc 
    A = dlmread('YOUR DATAS'); %Firstly add to path 
    plot(A(:,1),A(:,2)) %Plotting the graph 
    hold on 

    min_peak = input('Just write a value that is higher than minimum peak values: '); 
%This value must be between requested peaks and non-requested peaks (you can see this in graph) 
    [yval, yval_i] = findpeaks(A(:,2),'MinPeakHeight',min_peak); %Finding peaks 
    scatter(A(yval_i,1), yval); %Showing peaks 
    Beta = []; 
    xval = []; 
    for k = 1:size(yval_i,1) %Finding x values corresponding to y (peak) values 
     xval1 = A(yval_i(k),1); 
     xval = [xval xval1]; 
    end 
     Theta = xval/2; 

    for i = 1:size(yval,1) %Finding half of max. peak values 
     yval_i1 = yval_i(i,1); 
     while (yval(i,1))/2 < A(yval_i1+1,2) 
     yval_i1 = yval_i1+1; 
     end 

     yval_i2 = yval_i(i,1); 
     while (yval(i,1))/2 < A(yval_i2-1,2) 
     yval_i2 = yval_i2-1; 
     end 

     plot(A(yval_i2,1)*ones(size(A(:,2))), A(:,2)); 
     plot(A(yval_i1,1)*ones(size(A(:,2))), A(:,2)); 
    %  hold on 
    %  scatter(A(yval_i1,1),A(yval_i1,2)) 
    %  scatter(A(yval_i2,1),A(yval_i2,2)) 
     B = abs(A(yval_i1,1)-A(yval_i2,1)); 
     Beta = [Beta B]; 
    end 
    Beta 

    K = 0.94; 
    Lambda = 1.5406e-10; 

    To = []; 
    for j = 1:size(Beta,2) 
    To1 = (K*Lambda)/(Beta(j)*cos(Theta(j))); 
    To = [To To1]; 
    end 
    To = abs(To) 

Answer 2

这里是工作的例子，假设你有一个正态分布。我在一个Jupyter控制台中运行它，所以如果你不这样做，你必须跳过“魔术线”（%matplotlib notebook）并在最后加上plt.show()。

%matplotlib notebook 
import matplotlib.pyplot as plt 
from scipy.optimize import curve_fit 
import numpy as np 

numb = 500        # data size 
fwhm_in = 3        # set FWHM for the artificial data 
sigma = fwhm_in/2/np.sqrt(2*np.log(2)) # calculate sigma 
xval = np.linspace(-10, 10, numb)   # calculate x and y values using the formula from Wikipedia (see link in question) 
yval = (sigma*np.sqrt(2*np.pi))**(-1)*np.exp(-(xval)**2/(2*sigma**2))+np.random.normal(0, 0.03, numb) 

def fitFunc(x, x0, sigm):     # this defines the fit-function 
    return (sigm*np.sqrt(2*np.pi))**(-1)*np.exp(-(x-x0)**2/(2*sigm**2)) 

guess = (0.5, 2)       # tell the code with which values it should start the iteration. Close but not equal to the real values 
fitParams, fitCovariance = curve_fit(fitFunc, xval, yval, guess) # do the actual fit 
print(fitParams) 

print('FWHM_calc = {:.3f}'.format(fwhm_in)) 
fwhm_fit = 2*fitParams[1]*np.sqrt(2*np.log(2)) # calculate the FWHM from the fitted sigma (= fitParams[1], since fitParams[0] is the offset x0) 
print('FWHM_fit = {:.3f}'.format(fwhm_fit)) 

plt.plot(xval,yval, 'r.', label='data') 
plt.plot(xval, fitFunc(xval, fitParams[0], fitParams[1]), 'k-', label='fit', linewidth = 3) 

plt.grid(True) 
plt.legend() 
ax = plt.gca() 
ax.axvline(fwhm_fit/2, color='b') 
ax.axvline(-fwhm_fit/2, color='b') 

Answer 3

我分享这是一个答案，但意见不提供我上传图片，以澄清问题抱歉。所以我用下面的图片说明了问题（顺便说一下，我不能编辑或删除我不情愿地写的评论）。 我，当我运行代码此图

这个图表说明了什么我正在寻找