Prof's Bonus Question for XRay Diffraction

[NeedPhysHelp8]

Homework Statement

Hi guys, so here's a bonus question on my assignment for materials course.

" A metallic crystal is irradiated with Xrays of a fixed wavelength. Bragg reflection is observed at angles of 23 and 73.5 degrees, as well as another angle phi which is intermediate between the above two values. Assuming the same set of crystal planes (separation d) are responsible for scattering in all three cases, calculate a possible value of phi in degrees. Are there other possible values for phi apart from the value that you found?

Homework Equations

<br /> 2dsin(\theta)= m \lambda<br />
i think this is the only equation we need for Bragg peaks because we didn't learn anything else

The Attempt at a Solution

k so i figured if the wavelenth, distance are constant then only m and \theta are changing so i got:

<br /> m1 = 2dsin23/ \lambda = 0.781 d/ \lamda <br /> and
m2= 2dsin73.5/ \lambda = 1.918 d/ \lambda<br />

I have absolutely no clue how to go from here? any help is much appreciated

 

[Maxim Zh]

The complete solution was deleted (it's not allowed). So if you didn't see it:

You can get m_1 / m_2 from the two last equations.
Then you should select a pair of integers which fit this ratio.
Any integer number between them will give you \varphi.

 

[NeedPhysHelp8]

Ok thanks i'll give that a try, and no I didn't see the complete solution. I'll report back when I get it

 

[NeedPhysHelp8]

so here's what i got:

<br /> \frac{m_{1}}{m_{2}} = 0.407<br />
from here I concluded that that ratio is approximately m_{1} = 2 and m_{2} = 5
so the possible values for m_{\phi}= 3 and 4
Then using m_{1}=2 I plugged that back into first equation above and got \frac{d}{\lambda} = 2.56 so \frac{\lambda}{d} = 0.391
So using m_{\phi}= 3 I got \phi = 35.9 and using m_{\phi}= 4 I got \phi = 51.4
And these values work great! Thanks a lot