Bragg angle and X-ray Radiation

[cutecarebear]

Homework Statement

In an experiment with X-ray radiation, you use a continuous radiation from an X-ray pipe. What is the highest energy of the X-ray radiation you can get if the pipe's acceleration voltage is 40 kV? If we use radiation with double the wavelength of the maximum energy we get enough intensity to do a Bragg scatter experiement with a crystal. We get constructive interference at an angle of 30 degrees (Bragg angle). What's the jitter constant (d) in the crystal?

Homework Equations

We have the Bragg equation:

2dsin(Θ)=λm

and the cutoff Equation

λmin= hc/eV

The Attempt at a Solution

Due to the conservation of energy the maximum energy of the radiation is 40 KeV. The cutoff wavelength is thus 31 pm. A photon with energy of at most 40 keV has wavelength of at least 31 pm.

Then, plugging into the Bragg formula we get:

2dsin30= 62 pm*m
2d(1/2)=62 pm * m
d= 62 pm*m

Is this correct? If anyone can give any advice it would be much appreciated! Thanks in advance.

 

[mark726]

Your solution is mostly correct, but there are a few things to note. The maximum energy of the X-ray radiation is actually 40 keV, not 40 kV. The Bragg equation should also be written as:

2dsin(Θ)=nλ

Where n is the order of diffraction (in this case, n = 1). Additionally, the Bragg angle should be in radians, so it should be:

sin(Θ) = λ/(2d)

Therefore, the Bragg angle would be:

Θ = sin^-1(31 pm*m/(2*62 pm*m)) = 0.25 radians = 14.3 degrees

To find the jitter constant, we can rearrange the Bragg equation to solve for d:

d = λ/(2sin(Θ))

Plugging in the values, we get:

d = (62 pm*m)/(2sin(14.3 degrees)) = 223 pm*m

So the jitter constant would be 223 pm*m. Note that this is just an approximation, as the actual value would depend on the specific crystal used in the experiment.