Finding Bragg-Plane Spacing for a Crystal in Diffract-Meter

[burgerkin]

Homework Statement

\lambda=0.34nm , Xray shines through a crystal in diffract-meter.

A 1st-order peek occurs at 38.2°. Find the corresponding Bragg-plane spacing for the crystal

Homework Equations

2d sin\theta=m \lambda

The Attempt at a Solution

I tried to use \theta as 38.2 to calculate and got the wrong answer, so I realized that the angle \theta is not the same as 38.2. \theta should be the incident angle the ray makes with the crystal surface.

Then I tried again using \theta= 90+38.2 degree, and I still got the wrong answer which is 0.216 nm.

Please help me out! thanks in advance.

 

[burgerkin]

Anybody please help!

 

[Gregg]

What is the right answer?

 

[burgerkin]

I do not know the answer. If I knew, it would give me some clues maybe.

 

[Gregg]

Well shouldn't the answer just be

d= \frac{m \lambda}{2 \sin \theta}

Just as in your equation?

How do you know you got the wrong answer?

 

[burgerkin]

It is online problem, I input the answer and got instant evaluation. I just used the formula above and got a wrong answer.

 

[burgerkin]

anyone?? please help!

 

[ehild]

Theta in the Bragg equation means the angle the reflected ray encloses with the crystal plane. 38.2° is the angle between the direction of the original ray with the reflected one. The incident ray is also at angle theta with the crystal plane. So what is the value of theta?

ehild

 

[burgerkin]

thanks so much!