8.6 Man-made X-rays
295
emitters. But if we can make the emitters emit together, the light will be coherent,
and also we will be able to produce a far more intense light. (If one emitter produces
radiation with wave amplitude E 1 , the incoherent intensity from N emitters will be
proportional to NE 2
1 , while the coherent intensity will be proportional to N 2 E 2
1 .)
Lasers have coherent light because the emitters are stimulated to emit just as waves
pass by. (See Sect. 7.23.) If we try to make an X-ray laser in the same fashion as
a visible-light laser, we find that the X-rays are too strongly absorbed in the base
material to get a lasting pulse. However, by producing bunched charges in a vacuum,
accelerating them, and then back scattering light from a visible-light laser, the backscattered light can be made to be Doppler-shifted into the X-ray bands. Because
the incoming light was coherent, the back scattered light will also have coherence.
These source of X-rays are called ‘free electron lasers’, although the X-ray radiation
is not directly amplified.
Because the X-rays from free electron lasers are coherent and can be made with
high intensities, the imaging of proteins and other molecules need no longer be
restricted to those materials which can be ordered at the molecular level, such as
in a crystal. Images of individual nanoscale particles in dynamical setting are now
possible. But there is a caveat: With the needed intensities, and because individual
X-ray photons transfer keV energies, each ‘picture’ you might capture effectively
destroys the subject! However, you can learn a lot about what structures WERE
there and how they might have been associating.
An estimation of the intensities you might require to illuminate a macromolecule
follows: we will probably want a resolution of 1 nm. For a molecule of area A =
1000 nm 2 , that means we will want at least 1000 X-ray photons to be scattered or
absorbed. The chance that a particle is scattered by a scattering center is measured
by its ‘cross section’, σ . The cross section is the effective area around a scattering
center perpendicular to the incoming beam such that if the scattered particle enters
that area, it will be scattered into a given range of angles or it will be absorbed.
Since K and L shell electrons resonate with X-ray photons for the high Z atoms,
their atoms will be good scatterers as well as absorbers of X-rays. The probability
that an X-ray photon gets scattered or absorbed will be about σ/A. The probability
that N photons get scattered will be about 1 − (1 − σ/A) N ≈ 1 − exp (−Nσ/A).
This number should get close to one by selecting N sufficiently large. This means
we will want an intensity bigger than about I ≈ Nhf /σ σt for an X-ray pulse of
duration t. For carbon atoms, the cross section is of the order of half a kilobarn in
the X-ray region, i.e. about 0.5 × 10 −7 nm 2 . This makes N ≈ A/σ > 10 10 for our
molecule with carbon as the principal scatterer/absorber.
8.6.5 Penetration of X-rays Through the Body
Because the incremental energy loss of an X-ray beam must be proportional to
the incremental distance of penetration, the intensity of the radiation behaves with
penetration distance as
295
emitters. But if we can make the emitters emit together, the light will be coherent,
and also we will be able to produce a far more intense light. (If one emitter produces
radiation with wave amplitude E 1 , the incoherent intensity from N emitters will be
proportional to NE 2
1 , while the coherent intensity will be proportional to N 2 E 2
1 .)
Lasers have coherent light because the emitters are stimulated to emit just as waves
pass by. (See Sect. 7.23.) If we try to make an X-ray laser in the same fashion as
a visible-light laser, we find that the X-rays are too strongly absorbed in the base
material to get a lasting pulse. However, by producing bunched charges in a vacuum,
accelerating them, and then back scattering light from a visible-light laser, the backscattered light can be made to be Doppler-shifted into the X-ray bands. Because
the incoming light was coherent, the back scattered light will also have coherence.
These source of X-rays are called ‘free electron lasers’, although the X-ray radiation
is not directly amplified.
Because the X-rays from free electron lasers are coherent and can be made with
high intensities, the imaging of proteins and other molecules need no longer be
restricted to those materials which can be ordered at the molecular level, such as
in a crystal. Images of individual nanoscale particles in dynamical setting are now
possible. But there is a caveat: With the needed intensities, and because individual
X-ray photons transfer keV energies, each ‘picture’ you might capture effectively
destroys the subject! However, you can learn a lot about what structures WERE
there and how they might have been associating.
An estimation of the intensities you might require to illuminate a macromolecule
follows: we will probably want a resolution of 1 nm. For a molecule of area A =
1000 nm 2 , that means we will want at least 1000 X-ray photons to be scattered or
absorbed. The chance that a particle is scattered by a scattering center is measured
by its ‘cross section’, σ . The cross section is the effective area around a scattering
center perpendicular to the incoming beam such that if the scattered particle enters
that area, it will be scattered into a given range of angles or it will be absorbed.
Since K and L shell electrons resonate with X-ray photons for the high Z atoms,
their atoms will be good scatterers as well as absorbers of X-rays. The probability
that an X-ray photon gets scattered or absorbed will be about σ/A. The probability
that N photons get scattered will be about 1 − (1 − σ/A) N ≈ 1 − exp (−Nσ/A).
This number should get close to one by selecting N sufficiently large. This means
we will want an intensity bigger than about I ≈ Nhf /σ σt for an X-ray pulse of
duration t. For carbon atoms, the cross section is of the order of half a kilobarn in
the X-ray region, i.e. about 0.5 × 10 −7 nm 2 . This makes N ≈ A/σ > 10 10 for our
molecule with carbon as the principal scatterer/absorber.
8.6.5 Penetration of X-rays Through the Body
Because the incremental energy loss of an X-ray beam must be proportional to
the incremental distance of penetration, the intensity of the radiation behaves with
penetration distance as