SME
University Physics III: Waves, Optics & Modern · Problem type
An element’s Kα X-ray
Use this for “Estimate the Kα X-ray energy and wavelength of copper (Z = 29).”
Diagram
λ = 0.1551 nm: X-rays.
c = λ × f
299,800,000 m/s = 0.1551 nm × 1.9334 × 10¹⁸ Hz
Atomic number Z = 29
X-ray energy E = 7,997 eV
This is an example. Type your own values to start a new problem.
E (
)
=
13.6
eV
×
(
−
1)
2
×
1
−
1
4
λ
Wavelength
calculated · nm
f
Frequency
worked out (not typed) · Hz
Formulas
E
=
10.2
×
(
Z
− 1)
2
7,997
=
10.2
×
(29 − 1)
2
λ
=
1,240
E
0.1551
=
1,240
7,997
f
=
2.998 ×
10
8
λ
×
10
-9
1.9334 ×
10
18
=
2.998 ×
10
8
0.1551 ×
10
-9
Assumptions
•
Kα is an electron dropping from the L shell (n = 2) into a hole in the K shell (n = 1).
•
The other 1s electron screens one unit of charge, so the jumping electron feels about Z − 1: E = 13.6(Z − 1)²(1 − 1 ÷ 4) eV.
•
hc = 1240 eV·nm and c = 2.998 × 10⁸ m/s. The estimate is within a few percent from about sodium to silver (Z ≈ 11–47); heavier atoms come out up to 15% low.
Step-by-step
WE KNOW
Z = 29
FIND
X-ray energy (E), wavelength (λ), frequency (f)
Step 1 · Find X-ray energy (E)
Use E = 10.2 × (Z − 1)²
Hydrogen’s
2
→
1
gap,
13.6
×
(1
−
1
4
)
=
10.2
eV,
times
the
screened
charge
squared.
E
=
10.2
×
(29 − 1)
2
E
=
10.2
×
28
2
E
=
10.2
×
784
E
=
7,997
eV
Step 2 · Find wavelength (λ)
Use λ = 1,240 ÷ E
hc
=
1240
eV·nm
over
the
photon’s
energy
in
eV
gives
λ
in
nm.
λ
=
1,240
7,997
λ
=
0.15506
nm
→
X-rays.
Step 3 · Find frequency (f)
Use f = 2.998 × 10⁸ ÷ (λ × 10⁻⁹)
The speed of light over the wavelength in metres is the number of waves passing each second.
f
=
2.998 ×
10
8
0.15506 ×
10
-9
f
=
1.9334 ×
10
18
Hz
Check
Put the numbers back in. Both sides should match.
7,997
=
10.2
×
(29 − 1)
2
✓
0.15506
=
1,240
7,997
✓
1.9334 ×
10
18
=
2.998 ×
10
8
0.15506 ×
10
-9
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.