Physics and Engineering of Radiation Detection

2.3. Interaction of Photons with Matter 87

Example:

Calculate the wavelength below which it would be impossible for photons to

ionize hydrogen atoms. The first ionization potential for hydrogen is 13.6eV.

Solution:

The minimum energy needed to ionize an atom is equal to the binding energy

of the most loosely bound electron. Since for hydrogen this energy is 13.6eV,

for photoelectric effect to be possible we must have

Eγ ≥ 13. 6 eV

⇒

hc

λmax

≥ (13.6)

(

1. 602 × 10 −^19

)

J

⇒λmax ≤

hc

(13.6) (1. 602 × 10 −^19 )

m

≤

(

6. 625 × 10 −^34

)(

2. 99 × 108

)

(13.6) (1. 602 × 10 −^19 )

≤ 9. 09 × 10 −^8 m=90. 9 nm

Hence a photon beam with a wavelength greater than 90.9nmwill not be able

to ionize hydrogen atoms no matter how high its intensity is.

A.2 ComptonScattering

Compton scattering refers to theinelasticscattering of photons from free or loosely
bound electrons which are at rest. Since the electron is almost free, it may also get
scattered as a result of the collision.
Compton scattering was first discovered and studied by Compton in 1923. During
an scattering experiment he found out that the wavelength of the scattered light was
different from that of the incident light. He successfully explained this phenomenon
by considering light to consist of quantized wave packets or photons.
Fig.2.3.4 shows this process for a bound electron. The reader may recall that the
binding energies of lowZelements are on the order of a few hundredeV, while the
γ-ray sources used in laboratories have energies in the range of hundreds ofkeV.
Therefore the bound electron can be consideredalmost freeandat restwith respect
to incident photons. In general, for orbital electrons, the Compton effect is more
probable than photoelectric effect if the energy of the incident photon is higher than
the binding energy of the innermost electron in the target atom.
Simple energy and linear momentum conservation laws can be used to derive the
relation between wavelengths of incident and scattered photons (see example below),

λ=λ 0 +

h

m 0 c

[1−cosθ]. (2.3.10)

Hereλ 0 andλrepresent wavelengths of incident and scattered photons respectively.
m 0 is the rest energy of electron andθis the angle between incident and scattered
photons (see Fig.2.3.4).