In Exercise 3.3 you were asked to find a numerical way of measuring the width of
ID: 3144037 • Letter: I
Question
In Exercise 3.3 you were asked to find a numerical way of measuring the width of the distributions. One commonly used method is to find the "full width at half the maximum" (FWHM). To find this you determine where the number of data is one-half of the value of the maximum, i.e. where N(x) = A/2. There will be two such points for a bell shaped curve. Then the FWHM is the difference between the right hand side value and the left hand side value of x. For a Gaussian distribution what is the mathematical relationship between the FWHM and the standard deviation?
Explanation / Answer
Hello
I know that the full -width-half maximum of the peak must be in radius ,but when I measured the FWHM ,it is in 2theta .
The scherror formula uses the FWHM of the peak in THETA and not in 2THETA
Do I need to half it into the equation?this is important because doing so doubles the request you get the crystal size
EWHM (°2theta)=38.169
Lambda=0.15406nm
D=(0.9*0.15406)/(0.0819*(2pl/360)*cis(38.169/2)
In case FWHM is not divided D=102.6nm
In case the FWHM is divided D=205.6nm
Done.......