Bragg Diffraction Homework: Find Longest Wavelength for 2 Interference Maxima

[Skeptic.]

Homework Statement

A beam of 3.55keV X-rays is directed at a crystal. As the angle of incidence is increased from zero, a first strong interferece maximum is found when the beam makes an angle of 18.0^o with the planes of the crystal.

Calculated d=5.67x10^-10 from this (distance between adjacent planes)

(c) Find the longest wavelength for which two interference maxima would be produced.

Homework Equations

[/B]
$2d\sin \theta = n\lambda$

The Attempt at a Solution

[/B]
I set n=2, since we're looking for the second interference maxima, so then $d\sin\theta = \lambda$
I was confused by where theta came from here, but to find the maximum wavelength I thought I would set $\theta$ = 90, but this doesn't make physical sense to me.

 

[BvU]

Look up a picture. And wonder what to do with the 18^o in the problem description.
Randomly picking $\theta=\pi/4$ indeed doesn't make sense :)

 

[Skeptic.]

Where does $\theta = \pi /4$ come into it?
I chose $\theta = \pi /2$ to maximise $d\sin\theta$
And isn't the 18^o only applicable for the earlier part of the question, when you're effectively given $\lambda$ as $\lambda = hc/E$? As when the wavelength changes, so will the angle for the first maxima.
Part (c) tells us we're calculating a new value for $\lambda$

Not sure how well I explained myself there. Sorry if it's incomprehensible!

 

[ehild]

The Attempt at a Solution

[/B]
I set n=2, since we're looking for the second interference maxima, so then $d\sin\theta = \lambda$
I was confused by where theta came from here, but to find the maximum wavelength I thought I would set $\theta$ = 90, but this doesn't make physical sense to me.

It happens when the incident ray falls perpendicularly at the crystal plane and reflects exactly backwards. ( In the picture, the rays are shifted for clarity.)

 

[BvU]

Where does $\theta = \pi /4$ come into it?
I chose $\theta = \pi /2$ to maximise $d\sin\theta$

My mistake. Still 90 degrees was a random choice: apparently to maximize. But the thing to do is to make it 'fit'.

 

[nasu]

90 degrees is the maximum range of angles that can be measured by a diffractometer. This corresponds to back-reflection (see ehild drawing), or 2θ=180 degree. In this case the diffraction condition is 2d=nλ. If you want to have the second order at this maximum angle, then you have
d=λ. The first order will be at 2dsinθ=d or sinθ=1/2.
If λ is larger than d, you get the first order but not the second order. It will require an angle with sin>1.
If λ is larger than 2d you don't get any peak.

Of course, in practice the range of angles is less tan 0-180, with restrictions at both ends.