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1
Corrections to an Ideal Ion Chamber:
•
Given: The following is our idealized ion chamber: vents to air in the atmosphere
that is dry for the moment.
Collection
Volume, V
Vent to air
•
The ionization in the cavity is proportional to the numbers of gas atoms in the
cavity.
•
Use the ideal gas law. The number of molecules is proportional to the gas density:
n∝ρ.
PV = nRT → n ÷V → ρ ∝
P
T
Where,
-- P = gas pressure
-- V = gas volume
-- n = number of gas molecules
-- R = gas constant
-- T = gas temperature
•
An ion chamber (and its electrometer) get calibrated at a National Institute of
Standards and Technology (NIST) calibration lab (like we have !!) at 22oC and 760
torr (1 torr = 1 mmHg = 1/760 atm = 0.1333 kPa).
Tcal = (273+ 22)K = 295K
Pcal = 760torr
•
We must always relate the present gas density to that which occurred at
calibration; using the above, we have: {caution: opposite of both PTP in TG-51, and
Mcorr/Mraw in TG-21}{Ahmedin Jemal, 2013 #41}
Lecture 26 MP 501 Kissick 2016
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n Tcal P
295K
p(torr)
ρ
=
=
⋅
=
⋅
o
ρcal ncal T Pcal 273K + T ( C) 760torr
•
Now, let’s undo the ‘dry’ air assumption: humid air is less dense. The water vapor
contains a lot of hydrogen. The density of humid air uses the following:
ρh
295K
p (torr ) − 0.3783 pw
=
⋅
o
ρcal 273K + T ( C )
760torr
Where,
-- ρ h is the density of humid air.
-- pw is the partial pressure of water (torr) in the air.
•
pw , the partial pressure of water, is calculated by
(
)
torr  RH %

pw = 19.827torr + 1.21 T ( oC ) − 22o C o 
C  100

Where,
-- RH% is the relative humidity of the air.
•
One value for the density of air is ρ = 0.0012929 g / cc , recall, this is dry air at 0oC
and 760 torr. See Attix Eq. 12.16.
•
Other Humidity Effects: ( W / e and ‘mass stopping power’ )
•
The (W / e) h for humid air is less than (W / e) d for dry air (= 33.97J/C); it is easier
to ionize the hydrogen in water than the oxygen and nitrogen in the air.
•
The ratio of (W / e) h (W / e) d is shown below (lower curve), from Attix. Page 327:
Lecture 26 MP 501 Kissick 2016
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•
Recall that we find the dose, D, as
D=
•
Q
M
W

 e



This allows us to express the ratio of charge produced in humid air, Qh , to that in
dry air, Qd , as
Qh (Dh M h /(W / e) h )
=
Qd (Dd M d /(W / e) d )
•
Assuming Bragg-Gray theory applies: D ∝ Φ ( dT / ρdx) and also we know M = ρV :
Qh (Φ (dT / ρdx) h ρ hV /(W / e) h )
=
Qd (Φ(dT / ρdx) d ρ dV /(W / e) d )
Lecture 26 MP 501 Kissick 2016
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•
Note that the z/A dependence in the stopping power is quite different when
something has a lot of H in it. z/A for hydrogen-1 is 1.0.
•
The density of humid air is lower (as we all know from the weather)
•
Therefore, we can write, and summarize, the effects of humidity as follows:
•
Ion-Chamber Saturation and Recombination:
•
The dose in the ion chamber gas is proportional to the charge of one sign produced,
Q. Both signs exist and travel in opposite directions, past each other, on their way
to different electrodes. They can combine and neutralize each other.
+
v
+
•
Recombination reduces with increasing potential between electrodes.
•
Recombination also generally increases with density.
•
When the recombination does not change upon further increases in electrode
potential, the ion chamber is said to be saturated.
•
When the electrode potential is increased further, then the migrating charges can
cause subsequent ionizations (avalanches), and this process is used in proportional
counters and in Geiger-Muller counters.
Lecture 26 MP 501 Kissick 2016
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•
The charge collection efficiency, f, is the ratio of the charge collected, Q’, to the
charge produced, Q. From Attix, page 331:
•
There is also a recombination dependence on initial ion concentration. High LET
(linear energy transfer) particles like alphas produce so many ions that
recombination will be larger than for low LET particles like electrons.
•
We will focus on volumetric or general recombination – initially uniformly
distributed ion concentrations in the cavity.
•
The drift velocity of the ions across the chamber (mobility) is proportional to the
electric field strength, E {units are volts/m}:
-- positive ions:
v1 = k1 E
-- negative ions:
v2 = k 2 E
Where k1 and k2 are the mobilities of the positive and negative ions respectively:
{units are m2/(Vs) = (m/s)/(V/m)}
Lecture 26 MP 501 Kissick 2016
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•
Note that ‘mobility’ is a terminal velocity and not an acceleration, because it
assumes a constant drag – continuous bumping of each other as the ions try to
accelerate towards the electrodes.
•
With an electronegative gas, an electron attaches to a gas molecule, and this makes
the positive and negative ion mobilities about equal.
•
With a nonelectronegative gas like N2, CO2, H2, methane, and noble gases, the
electron does not attach, and so the very light electrons have a much higher
mobility {~103cm2/(Vs)}.
•
Consider the following situation: parallel plate chamber with an electronegative gas
in a continuous radiation field.
•
Just after the irradiation starts, the positive ion charge density, ρ1(x), {units are
C/m3}, looks like the following:
t = ∆t
ρ1(x)
t = 2∆t
+
-
t = 3∆t
+
-
+
-
2∆x
∆x
q∆t.
x
Positive ions created
between 0 and ∆t.
x
Positive ions created
between ∆t and 2∆t.
x
Positive ions created
between 2∆t and 3∆t.
Lecture 26 MP 501 Kissick 2016
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•
The distance the ions move between frames 1 and 2 is ∆x:
∆x = v1∆t = k1 E∆t
Where ∆t is the time between frames.
•
The ionization density rate is q {units are esu/(m3s)}, and the time it takes a
positive ion to cross the cavity from its creation at the opposite electrode is τ1:
τ 1 = d / v1 =
d
k1 E
•
There is an analogous relation for the negative ions, τ2.
•
At times greater than τ 1 and τ 2 , the steady state charge density is achieved. The
charge density of positive ions just next to the negative electrode is:
ρ1 ( x = d ) = q∆t
τ1
∆t
= qτ 1 =
qd
k1E
&
q=
ρ
t
-- And likewise for the negative ions, BUT if the mobility if different, then the
steady state charge density will be too!
•
In general, one might get a steady state charge distribution like the following:
units:
d
(esu / m s )m
esu
 qd 
 kE  = (m 2 / Vs)(V / m) = m 3
3
-- Note: highest
recombination
in the middle.
+
qd
k2 E
k1 < k 2
ρ2(x)
(- ions)
Also, the level
depends on the
mobility.
-
qd
k1 E
ρ1(x)
(+ ions)
x
Recombination highest
where both species are
present.
Lecture 26 MP 501 Kissick 2016
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•
The above was derived without recombination, but it’s an OK approximation if
recombination is small.
•
As mentioned, the recombination rate per volume, r(x), {units of C/(m3s)}, is
proportional to the charge densities of both species:
r ( x) =
α
e
ρ1 ( x) ρ 2 ( x)
Where α is a constant: the recombination constant, {units of m3/s}.
•
The total (integrated) recombination rate, R, is as follows:
R=
α
e
d
∫ ρ ( x) ρ
1
2
( x)dx
0
-- Inserting what we already have,
R=
•
α d  qd  x   qd 
αq 2
x 




⋅
1
−
dx
=
 



e ∫0  k1 E  d   k 2 E  d 
ek1k 2 E 2
0
6ek1k 2 E
Now we can use these equations for the collection efficiency, f. Note that the
charge produced per unit area and per unit time is qd. Therefore,
f = 1−
•
αq 2 d 3
{
(
x
)
⋅
(
d
−
x
)
}
dx
=
2
∫
d
R
αqd 2
= 1−
qd
6ek1k2 E 2
For a parallel plate geometry, the electric field, E, is related to the potential, P, by
E = P/d. Therefore,
1
f = 1− ξ 2
6
Where, we define ξ as:
ξ≡
d2 q
P
α
ek1k 2
=m
d2 q
P
and m = 36.7Vs1/2cm-1/2esu-1/2 for air at (one type of) STP (= 0oC, 760 torr).
Lecture 26 MP 501 Kissick 2016
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•
We actually should have considered recombination when giving the charge densities
… We actually overestimated the recombination.
•
Fix by integrating with the charge densities at the electrodes as:
fqd
fqd
&
k1 E
k2 E
1 2 2 Q'
f ξ =
6
Q
•
We would then get: ( R / qd ) = ( f 2ξ 2 / 6) and f = 1 −
•
However, this now underestimates the recombination, because we have not
considered the more complex shape of the charge densities – space charge
alteration of the electric field (see blue lines):
-- Note that we specify charge density at the electrodes by the potential:
d
k1 < k 2
Recombination affected
electric field magnitude
+
fqd 2
k 2V
ρ1(x)
(+ ions)
ρ2(x)
(- ions)
fqd 2
k1V
x
•
One can do this even better. Mie’s theory would give:
R 1 2
1
= fξ and f = 1 − fξ 2 (not quadradic this time), and this gives
qd 6
6
Attix Eqn. 12.26:
f =
•
In summary then:
1
1 + (ξ 2 / 6)
}
1 {0,1, 2} 2
f = 1− f
ξ
6
Exponent of
0: E const., no recomb., 1: E adjusted with recomb., 2: E const. with recomb.
Lecture 26 MP 501 Kissick 2016
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•
Let’s look at the same thing in a different way, …, following Johns & Cunningham,
Chapter 9:
•
First, about units and books/notes:
-- In our notes: r ( x ) =
α
e
ρ1 ( x) ρ 2 ( x) above, has units: [r ] =
C (recombined )
m3s
C
C
m3
with: [ρ ] = 3 , [e] =
, [α ] =
m
esu
s
-- In our notes and in Attix: [q ] =
esu
.
m3 s
-- In Johns & Cunningham: Q ⇔ ρ , Not the same as Attix Q !
•
Adopting Johns & Cunningham notation and letting ρ1 = ρ 2 = Q .
-- in less than, say, 1 ms, Q ≠ func(t ) .
-- Therefore,
R=
α
Q2
e
-- for longer times, replace their ‘r’ with dQ/dt {‘recombination loss rate density’},
dQ =
−α 2
C
Q dt , and recall: [Q] = 3 .
e
m
-- integrate to obtain:
α
dQ'
∫Q Q'2 = − ∫0 e dt '
0
Q
t
Q
1
−α
t
−
=
Q' Q0
e
α
Q
= 1−
Q0
e
Q
Q0
 Q0t ⇒ Q =
α 
 Q0 
1 +  Q0t
e
Lecture 26 MP 501 Kissick 2016
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•
Q
Q'
from earlier,
⇔
Q0
Q
One can identify:
f ⇔
-- Therefore,
α 
f = 1 −  Q0tf
e
-- Recall, we had:
1
f = 1 −   fξ 2
6
-- We can then identify:
ξ2
α 
⇔  Q0t
6
e
f = 1 − fξ 2 =
-- Given that:
… Going further …
α q
d4
dQ
and
= eq
e P k1k 2
dt
⋅
⋅
{Note that: E(electric field) = P(potential)/d(parallel plate separation)}
•
Note also the notation difference between
Johns & Cunningham:
[Q] =
Attix:
[Q] = C
C
m3
•
Back to an Attix-based formulation …
•
Let us find the charge produced in the following way (if ‘f’ not too far from 1):
1 Q
1
1 m2d 4q
=
≈1+ ξ 2 =1+
f Q'
6
6 P2
-- then, if the irradiation time, t, is much longer than τ:
Q = qVt →
Q
cQ
= 1+ 2
Q'
P
Lecture 26 MP 501 Kissick 2016
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Where, c ≡
m2d 4
, and dividing by Q gives the following linear equation:
6Vt
1 1 c
= +
Q' Q P 2
And let’s use this to reproduce the figure in Attix, page 335: (This is called a Jaffe
Plot)
1 1
c
= +
Q' Q P 2
slope = c
1 Q '2
1 Q'
1 Q '1
1Q
1 P2
2
1
1P
2
2
1P
-- Note that it is often easier to use a plot of 1/f and 1/Potential instead of the
following, because it would be a straight line normally ..
•
We can use this, extrapolating from two measurements, to get Q, the charge
produced ! This is also called the ‘saturation charge.’ It is written as Q0 in Johns &
Cunningham notation.
•
If the radiation is pulsed, then this steady state will likely not be reached: a typical
linac has 1 µs pulses that are 1 ms apart.
•
Recombination requires significant overlapping charge densities.
exploring confounding issues more closely:
•
In both TG-21 (IV.c and Fig. 4) and TG-51 (Eqs. 11 and 12), there is a ‘two-voltage’
technique to correct for recombination that relies on the above figure being linear.
For typical chambers, errors of < 0.5% occur because of that assumption.
It is worth
Lecture 26 MP 501 Kissick 2016
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•
For typical ion chambers, there are other factors such as prompt (‘initial’)
recombination and ionic diffusion that also can lead to charge reduction.
•
The ion chamber, plateau, region of the response to applied potential is also not
truly flat because of some charge multiplication.
•
In smaller chambers, like the A1SL, these effects can be more pronounced since
the region near walls, that has more varying electric fields, is a larger portion of
the total volume. Therefore, deviations from linearity, or a pure ion chamber type
response can be more pronounced in small or micro-chambers.
•
In C. Zankowski and E.B. Podgorsak, Med. Phys. 25 (1998) 908-15, The following
form (their eq. 14) is used to include these corrections (in our notation):




general
recombination

} 

1 1
α
β
= +
+
−γP}
exp{
2
1424
3
Q'  Q
P
P
{
 ch arg e
initial

 multiplication
recombination ,
ionic


diffusion


-- The effects look like the following (note different abscissa!), their fig. 2:
Note that V is the same as P for ‘potential,’ and charge multiplication at high V.
Lecture 26 MP 501 Kissick 2016
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•
Ionization, Excitation, and W / e :
•
One never directly measures dose with an ion chamber.
-- one measures ionizations only (neglecting excitations, but W / e has this implicitly
included, yet we see the complications with this too (i.e., humidity).
-- even when correcting for recombination, we are still not even directly measuring
charges produced.
•
So, (W / e) gas is not just the energy required to ionize a gas molecule: it is the
relation between energy deposited and the ionization produced. So, it is not really
the ionization potential, but rather a work function. See Attix, pgs. 340-341.
•
In other detector systems, it is reversed.
Scintillation detectors and
thermoluminescent detectors have the trouble of measuring just excitations, and
then need to relate that signal to the deposited dose.
•
To measure (W / e) gas , we can use a radioactive gas, and we need a large enough
chamber and thick enough walls to capture all of the charged particle kinetic
energy, except that which is released in bremsstrahlung:
N decays of
energy, T0
V
A
Where N is the number of charged particles produced.
W
 e
•
This device allows for 
•
What we find is that 
W
 e

N ⋅ T0 (1 − g )
 =
Q
 gas
 W
 ≠ 
e  e

 , but at least they are constant over energy.
α
Lecture 26 MP 501 Kissick 2016
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•
Now, let’s explore an Example Problem: Johns & Cunningham, page 333, #1:
-- Radiation pulse creates: Q0 = ρ1 = ρ 2 .
-- Given: α = 1.6 × 10 −6 cm 3 / s and Q0 = 20 pC / cm 3 .
Q 1 1
= , ,... {and plot it …}
Q0 2 4
-- How long until:
•
Solve by using: Q =
•
Input values:
t1/ 2
α 
1 +  Q0t
e
⇒t =
e
αQ0
 Q0


− 1 .
Q

1.6 × 10 −19 C
 1

=
− 1 = 5ms
−6
3
−12
3 
(1.6 × 10 cm / s ) ⋅ (20 × 10 C / cm )  1 / 2 
t1/ 4 =
•
Q0
1.6 × 10 −19 C
 1

− 1 = 15ms
−6
3
−12
3 
(1.6 × 10 cm / s ) ⋅ (20 × 10 C / cm )  1 / 4 
Plot it:
Log(Q)
100
10
If exponential,
would be straight line …
1
0
•
5
10
15
20
Time, ms
Time between pulses allows for charges to recombine!
Lecture 26 MP 501 Kissick 2016
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•
This is important because calibration protocols use a factor for recombination, such
as Pion = 1/f. We calibrate to Co-60 which is continuous versus our linacs that are
pulsed: From Attix, Page 338:
•
‘f’ is higher for a continuous beam! The ion transit time is about a ms. Therefore,
Co-60 has a negligible recombination correction.
•
This leads to the two voltage technique used in the protocols, and
 Λ + Λ d Λ g + Λi ⋅ Λd 
f = f i ⋅ f d ⋅ f g ~ 1 + i
+

V
V2


-- The expansion factors: Λ g , Λ i , Λ d relate to recombination. ‘i’ is for ‘initial,’ ‘d’ is
for ‘diffusion,’ and ‘g’ is for ‘general.’ The ‘general’ factor is dose rate dependent.
-- A quadratic dependence results from: Λ i ≈ Λ d ≈ 0 .
•
Next lecture: we start to synthesize all this information for clinical protocols !!
Lecture 26 MP 501 Kissick 2016