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RE: QUESTION regarding radiation theraopy using linear accelerator that produces a 6 MEV photon
Actually, 6 MeV accelerators are the minimum energy that is typically used
for breast cancer, or most other cancers for that matter.
The physics of high energy x-rays and the thickness of the body part to be
treated dictate what energy will be used.
Low energy (e.g., < 300 kVp) x-rays or gamma-rays have a limited ability to
penetrate tissue. Since most cancers are not superficial, low energy beams
are not used for most therapy.
Passage of high energy photons through tissue causes a build up effect.
That is, the maximum dose occurs not at the surface of the skin, but at some
depth. For Co-60 gamma rays this is about 0.5 cm, for 6 MVp x-rays it is
1.5 cm. This has the effect of sparing the top surface layers of the skin
(epidermis and parts of the dermis). These tissues are more radiation
sensitive than deeper structures. Hence, more radiation dose can be
delivered to the cancer before serious side effects occur.
In breast cancer, the entire breast is at risk for relapse. So, it is
important to uniformly treat the entire structure. Because of its
penetrating ability, a 6 MVp x-ray beam is able to uniformly irradiate the
entire breast when used in a parallel opposed configuration (two beams
delivered 180 degrees apart). Beams of a lower energy (e.g., Co-60) are
unable to penetrate deeply enough to provide such uniform coverage. For
some women with unusually large breasts even higher energies may be
necessary (e.g., 15 MVp).
Electrons are not used, by themselves, for breast cancer therapy because
they have a much more limited penetrating ability than x-rays.
-----Original Message-----
From: Ted de Castro [mailto:tdc@XRAYTED.COM]
Sent: Wednesday, April 10, 2002 7:27 AM
To: radsafe@list.vanderbilt.edu
Subject: Re: QUESTION regarding radiation theraopy using linear accelerator
that produces a 6 MEV photon
Well - actually not quite.
The accelerator accelerates electrons and does so all at the same energy
- it cannot work any other way. To get photons the electrons strike a
target and then produce a spectrum of photons.
The point was well made about 6 meV photons being a surprisingly high
for this therapy. I don't do this sort of work so I DON'T KNOW - BUT -
I would suspect that MAYBE they are using the electron beam at an energy
set so as to put the Bragg Peak at the depth of the cancer. I DO KNOW
that this is the method used when heavy ion accelerators are used for
breast therapies.
But then again - I don't know that this is not a generally distributed
cancer. If it were - then I would GUESS that the high energy would be
used in a lateral orientation in order to achieve a fairly flat dose
across the entire organ.
I was waiting to see if someone with more specific information would
weigh in on this subject before stepping in myself.
alstonchris@netscape.net wrote:
>
> A linac is basically a super-powered xray machine. It's not producing a "6
> MEV photon". It's producing a beam of a continuous range of energies, up
to a max E of 6 MEV. So, it's a spectrum like, in diagnostic xray language,
a 6 MVp beam, but it's accelerated by RF, not a voltage potential. The
effective E might be, nominally, 2 MEV. Hint: remember that what delivers
the *dose* is electrons.
>
> The RHH is not a great deal of help, in matters of radiological physics,
that's not what it's designed for. I highly recommend: Khan, Faiz M; The
Physics of Radiation Therapy; Williams & Wilkins. For a question like this,
the edition number is of no account, I should think.
>
> Cheers (and best wishes for your friend)
> cja
>
> "Sandy Perle" <sandyfl@EARTHLINK.NET> wrote:
> >I am forwarding this question for Michael Coogen. Please respond to
> >Michael directly at: michael_coogen@hq.dla.mil
> >
> >I have a very dear friend whose wife has been diagnosed with
> >adenocarcinoma of the breast. The radiation therapist is using a
> >linear accelerator that produces a 6 MEV photon. Looking at Figure
> >5.3, page 129, in the RHH, June 84 edition, which is for water
> >absorption (it's the closest graph that is comparative for tissue),
> >and using the I=Ioe-(u/p)(p)(t) equation, I calculate that 0.8 of the
> >photons pass right on through the breast. So I ask why use such high
> >energy photons when most of them are not absorbed. Then looking at
> >the same graph for Compton scattering, at 6 MEV, the scattering
> >calculates to be only 7%. At 180 centigrays (rads/day), I conclude
> >that they need the very high energy photon to minimize compton
> >scattering to the rest of her body. I also suspect that the cone
> >created by the scattered photons is less as the energy goes up. Am I
> >right with these assumptions? Can anyone shed some light on the type
> >of therapy?
> >
> >Michael Coogen
> >michael_coogen@hq.dla.mil
> >
> >-------------------------------------------------
> >Sandy Perle
> >Director, Technical
> >ICN Worldwide Dosimetry Service
> >ICN Plaza, 3300 Hyland Avenue
> >Costa Mesa, CA 92626
> >
> >Tel:(714) 545-0100 / (800) 548-5100 Extension 2306
> >Fax:(714) 668-3149
> >
> >E-Mail: sandyfl@earthlink.net
> >E-Mail: sperle@icnpharm.com
> >
> >Personal Website: http://sandy-travels.com
> >ICN Worldwide Dosimetry Website: http://www.dosimetry.com
> >
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