SME
Materials Science & Material Properties · Problem type
X-ray peak by Bragg’s law
Use this for “Copper’s cell edge is 0.3615 nm. At what angle 2θ does the (111) peak appear with 0.1542 nm X-rays?”
Diagram
The
(111)
plane
cuts
the
axes
at
a
h
,
a
k
and
a
l
(never,
for
a
0);
the
next
plane
is
d
=
a
÷
√
h
2
+
k
2
+
l
2
farther
on.
The wave off the next plane travels 2d sin θ farther: a whole number of wavelengths, nλ = 2d sin θ, and the two add.
This is an example. Type your own values to start a new problem.
λ (nm)
=
2
×
d (nm)
×
sin(
θ
°)
d (nm)
=
a (nm)
h
2
+
k
2
+
l
2
2θ
Detector angle
calculated · °
Formulas
d
=
a
÷
√
h
2
+
k
2
+
l
2
0.2087
=
0.3615
÷
√
1
2
+
1
2
+
1
2
λ
=
2
×
d
×
sin(
θ
°)
0.1542 = 2 × 0.2087 × sin(21.6791°)
2
θ
=
2
×
θ
43.3583 = 2 × 21.6791
Assumptions
•
A cubic cell: planes (hkl) are d = a ÷ √(h² + k² + l²) apart.
•
First-order peak (n = 1): the extra path 2d sin θ is one wavelength.
•
FCC metals show peaks only when h, k, l are all odd or all even.
Step-by-step
WE KNOW
a = 0.3615 nm, h = 1, k = 1, l = 1, λ = 0.1542 nm
FIND
plane spacing (d), Bragg angle (θ), detector angle (2θ)
Step 1 · Find plane spacing (d)
Use d = a ÷ √(h² + k² + l²)
In
a
cubic
cell
the
planes
(hkl)
are
a
÷
√
h
2
+
k
2
+
l
2
apart.
d
=
0.3615
÷
√
1
2
+
1
2
+
1
2
d
=
0.3615
÷
√
1 + 1 + 1
d
=
0.3615
÷
√
3
d
=
0.3615
1.732
d
=
0.2087
nm
Step 2 · Find Bragg angle (θ)
Use λ = 2 × d × sin(θ°)
The
angle
whose
sine
is
λ
2
d
;
there
is
no
peak
when
λ
is
more
than
2
d
.
θ
=
sin⁻¹(
λ
2 ×
d
)
θ
=
sin⁻¹(
0.1542
2 × 0.2087
)
θ
=
21.67913°
Step 3 · Find detector angle (2θ)
Use 2θ = 2 × θ
The
detector
reads
2
θ
,
the
angle
between
the
beam
in
and
out.
2
θ
=
2
×
21.67913
2
θ
=
43.3583°
Check
Put the numbers back in. Both sides should match.
0.2087
=
0.3615
÷
√
1
2
+
1
2
+
1
2
✓
0.1542 = 2 × 0.2087 × sin(21.67913°) ✓
43.3583 = 2 × 21.67913 ✓
Topic check
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