SME
Mineralogy · Problem type
A cubic plane’s spacing and its peak
Use this for “Halite is cubic with a = 5.640 Å. Find the spacing of its (200) planes and their 2θ peak with copper Kα.”
Diagram
The
(200)
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.
This is an example. Type your own values to start a new problem.
d (Å)
=
a (Å)
÷
h
2
+
k
2
+
l
2
λ
X-ray wavelength
example
θ
Bragg angle
worked out (not typed) · °
2θ
Detector angle
calculated · °
Formulas
d
=
a
÷
√
h
2
+
k
2
+
l
2
2.82
=
5.64
÷
√
2
2
+
0
2
+
0
2
λ
=
2
×
d
×
sin(
θ
)
1.5406 = 2 × 2.82 × sin(15.8522°)
θ
=
2
θ
2
15.8522
=
31.7043
2
Assumptions
•
Cubic cells only: the planes (hkl) are a ÷ √(h² + k² + l²) apart; other systems have their own formulas.
•
First order (n = 1): λ = 2d sin θ, and the detector reads 2θ.
•
Copper Kα X-rays have λ = 1.5406 Å.
Step-by-step
WE KNOW
a = 5.64 Å, h = 2, k = 0, l = 0, λ = 1.5406 Å
FIND
plane spacing (d), Bragg angle (θ), detector angle (2θ)
Step 1 · Find plane spacing (d)
Use d = a ÷ √(h² + k² + l²)
Higher indices cut the cell more often, so their planes sit closer together.
d
=
5.64
÷
√
2
2
+
0
2
+
0
2
d
=
5.64
÷
√
4
d
=
5.64
2
d
=
2.82
Å
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
that
is
more
than
1.
θ
=
sin⁻¹(
λ
2 ×
d
)
θ
=
sin⁻¹(
1.5406
2 × 2.82
)
θ
=
15.85216°
Step 3 · Find detector angle (2θ)
Use θ = (2θ) ÷ 2
The
beam
turns
through
twice
the
Bragg
angle
θ
.
2
θ
=
2
×
θ
2
θ
=
2
×
15.85216
2
θ
=
31.7043°
Check
Put the numbers back in. Both sides should match.
2.82
=
5.64
÷
√
2
2
+
0
2
+
0
2
✓
1.5406 = 2 × 2.82 × sin(15.85216°) ✓
15.85216
=
31.70431
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Crystallography
Main lesson
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