PHY 217
Q1.
Short Practice Problems- Answers Spring 2000 Set
Consider a thin target used to generate X photons by bremsstrahlung. Suppose the incoming electrons
kinetic energy at the face of such a target is tripled. In this case what will occur to the intensity
density functions value ?
(it reduces by x(1/3) at all energy values)
Q2.
Consider the plot of an intensity density function, I(e), for a thick target generated spectrum which has
an intercept on the I() axis such that I(0) = 500 (in suitable units). What will be the value of I(0) in the
same units when only the tube current is tripled?
(triples to 1500)
Q3.
Suppose I(0) = 250 m-2.s-1 and I(60) = 0, for a Kramer intensity density spectrum plot of a THICK
targets generated X -photons. Then the value of total X ray intensity from the target is, (assuming
photon energy measured in keV) :
(250 x 30 = area of triangle = 7500 keVm-2s-1)
Q5.
For a (Kramer) intensity density spectrum for a THIN targets generated X-photons, where the Duane
Hunt limit is 120 keV, and given that I(80) = 350 m-2s-1, then the intensity due to photons between
10keV and 22 keV is?
(Use fact of constant value for I(), then  intensity = (350/m2*s)*(12keV) = 4,200 keV/(m2s))
Q6.
A Kramer THICK target generated spectrum plot of I() is such that I(0) = 1012 m-2.s-1 and I(70) = 0.
Then the value of X ray intensity from the target due only to photons between 49 and 50 keV is ?
(Use integral from 49 to 50 of (10 12/70)*(70-E) )=2.928571429*1011)
Q7.
In the above question (Q6) case, what is the (approximate) total number of photons emitted between 49
and 50 keV in 1 second?
( Use photon flux density = intensity density/E, then get integral of photon flux density from E=49 to
E=50 => 5.916*109~ 6*109 , or see that I(49.5)/49.5= [(1012/70)*(70-49.5)]/49.5 = 5.916*109 ~
6*109)
Q8.
For an X-photon beam comprized of only two photon energy values, the fraction of beam intensity
transmitted through a depth x of material is given by 0.8 exp(-0.15.x) + c.exp(-0.4x).
Then the value of the constant c is:
(c = 0.2 : Use sum of two constants at x = 0 is 1)

Q
Homogeneous material is irradiated whereby the photon fluence from an X ray tube is 1020 photons per
square metre which have aneffective photon energy of 65keVcorresponding to a linear attenuation
coefficient of 46/metre. The detector is 70% efficient, the materials thickness is 40 cm and the
secondary to primary fluence is 6. Then the detector energy fluence at the centre of the detector is:
10
20.
(Use (x,y) = (10^20) * (1+6)*(65000eV*(1.6*10^(-19)(J/eV)))*(70%)*exp(-(46).(0.4)) =
19 .
46. 0.4
.
7 65000 . 10
( 0.7 ). e
= 0.03251554 )=0.03251554 J/m2
Q11.
Asuming only the secondary to primary ratio changes, then for an increase in secondary to primary
ratio from 3 to 5 we expect a change factor in image signal at the detectors centre of ?
(Use same eqn as above (general) see that the factor concerned is (1+R), then in first case, (1+3)=4,
and in second case (1+5)=6, hence increase factor is (6/4)= 3/2).
Q12.
At a certain exposure rate, the X radiation intensity onto the input phosphor of an Image Intensifier is
10-5 Jm-2s-1. The effective photon energy is 44 keV. Then the the quantum SNR value corresponding to
a pixel on the TV screen representing a 2mm x 2mm square of the input phosphor and over the 0.2
seconds of eye integration time is:
(Use Ncollected at II sqmm in 0.2sec = [(10-5)/(4.4*104*1.6*10^-19)]*[4*10-6*0.2]
10
4.
5
.
.
= 4.4 10 1.6 10
Q13.
19
. 4. 10 6. 0.2 = 1.136363636 10 3
, so sqrt N= ~33.7 ~ 34)
Suppose contrast, C, of an entity yielding an air Kerma of K2 is defined by C = ln(K2/K1) where K1 is th
surrounding air Kerma value. Taking a simple geometry of a cube of ice, suppose there is a small non
attenuating cubic cavity of sides 2mm, within the larger ice cube . The linear attenuation coefficient of
the ice is 32/metre. The contrast of the non attenuating cavity is, (disregarding scatter effects):
(Use C = +x = + 32*(2/1000) for this case= 64/1000)
Q14.
Suppose now the geometry of Q13 above and the definition of contrast used there, but that the small
cube is an 'embedded object' that has an attenuation coefficient of 22/metre.
The contrast of the embedded object is (disregarding scatter effects):
(Use  =  = 32-22 , and C = -*x2 = = 20/1000 = -0.02)
Q15.
The simple radiographic contrast equation ( which is based upon a definition of contrast, C, of the form
C= (), also takes 'scatter' X radiation into account. If the ratio of scatter to primary energy
fluence is increased from a value of three to six, this will lead to a multiplacative factor change in C of
? :
(C prop' to 1/(1+R) => case1 is C1 = k/4, case2, C2 = k/7. Thus C1*f = C2, => f = C2/C1 = 4/7)
Q15.
Suppose that the photon fluence increases fivefold at the entrance to a radiographed part. Then in this
case the signal to noise ratio.....?.....:
(Increases by factor of sqrt(5) ie f ~ x (2.24))
Q16.
Suppose the cross sectional area of a target we are trying to visualize is doubled, then the signal to
noise ratio......?.......:
(Changes by a multiplacative factor of sqrt 2 ~ x 1.414)
Q18.
Suppose a small liver tumour is imaged (by CT, say) and it is a 0.6 cm3 cube. By what factor would the
minimum skin absorbed dose to the patient change in order to be able to detect a 0.1 cm sided cubic
mass of the same material?
(Use Dmin prop to 1/x^4, so get Dmin1=k/(0.6)^4, Dmin2= k/(0.1)^4=> f= 6^4 = x 1,296)
Q19.
Supposing that a liver tumour as is imaged but contrast agent given increases the difference between
the tumours linear attenuation coefficient and and the surrounding liver attenuation coefficient
fivefold. Then the minimum skin absorbed dose to the patient to be able to detect the tumour changes
by what factor?
( 1/25, <=Use fact Dmin prop' to 1/(square of  ))
Q20.
A particular radiograph has value of 3mm for each individual unsharpness (geometric, absorption,
movement, screen). Then the total unsharpness (in mm) is:
(Use sq rt of all the squares => 6)
Q21.
An imaging system setup has a screen unsharpness of 1.5mm, and a magnification of 3.5.
Then the standardized screen unsharpness is?
(Use Ws = F/m, so => 1.5/3.5= 0.3/0.7 = 3/7)
Q22.
Suppose no absorption nor movement unsharpness in a system, with standardized screen unsharpness
0.5mm and standardized geometric unsharpness 1mm. Then the total standardized unsharpness (in
mm) is:
CHECK
(Use Wt =k. sqrt (Ws + Wg) => Wt=1*sqrt{
Q23.
?+? } (, noting k = 1 mm2))
A fluorescent screen has an absorptive efficiency of 60%, an intrinsic sensitivity efficiency of 75%, and
a screen sensitivity efficiency of95%. Then the (approximate) overall efficiency is:
(Use Etot = Ea*Ei*Escr=> 0.6*0.75*0.95= 0.4275~ 43%)
Q24.
The intensity of visible light passing through a processed (blackened) film decreases with the thickness
of the emulsion. For a given number of silver specks per unit volume of the emulsion, there is a definite
relationship between the rate of change of the light intensity , cross sectional speck area,
film thickness, and number of specks per unit area of film.
Suppose that the cross sectional area of speck is doubled, the number of specks per unit area is
quadrupled and the film thickness halved. Then the loss of intensity per unit distance is:
(Use dI/dy = -I.c.g/t => x 16 fold )
Q25.
The 'variable kVp rule' is sometimes stated as "add 5% of kVp per centimetre thicker part".
If we accept this rule, write down a valid Euler scheme for this rule?
(
Vj+1= 1.05*Vj
)
Q27.
A model of image density, D, consistent with the 'traditional' quoted rules of X ray 'exposure' control
assumes what relationship between D and the tube voltage, V ?
(D prop to V^5)
Q28.
A model of image density, D, consistent with the 'traditional' quoted rules of X ray 'exposure' control
assumes what relationship between logeD and the part thickness, x ?
( D is prop' to exp(-u.x) or equivalent)
Q29.
A film emulsion has 3000 un sensitized grains per mm2 (of silver bromide) prior to any irradiation.
Given that the facial area of a grain is 2m x 2 m, then when 70% of the grains present are sensitized,
the rate of change of sensitized grains (silver specks) with photon fluence is equal to:
(Use dg/dN = (G-g)*b , then g = 2100, G-g = 900, convert area to 4*10-12 m2 => 900*4*10^-12
= 3600*10^-12 = 36*10-10 sensitized grains.m2.photon-1. )
Q30.
The maximum density on a film possible is 2.7, and the typical size of a film speck is 3m x 3m.
Then Nuttings law predicts the number of unsensitized grains per unit area to be (approx):
(Use Dm = 0.4343*G*sigma, => 2.7 = 0.4343*G*(9x10 -12 m2), => G ~ 7*1011 )
Q31.
A film with a density difference of 0.7 in a region of density 3.0 is developed to give a 'gamma'
for the film of 3.5. This means that the exposure contrast required to produce this film is
approximately:
a.
~ 0.039
b.
~ 0.46
c.
~ 1 Ckg-1.
d.
~ 4 C kg-1.
e.
~ 0.25
***
(Use CB = -Gamma*Cx => Cx = (-2.303)*(0.7)*(-1/3.5) ~+0.4606 , the density value is red herring)
Q32.
Consider a collexion of four equal sized voxels, each voxel having the following slice representation to
a fan beam and nominal values of linear attenuation coefficient :
1
2
3
7
Then the full list of ray sums usable in computation is :
a.
3, 4, 5, 8, 9, 10
***
b.
3, 4, 9, 10
c.
3, 10
d.
2, 3, 14, 21
e.
2, 3, 6, 7, 14, 21
(use all angles => 6 possible SUMS)
Q33.
A tissue has a linear attenuation coefficient of 33 m-1. For the same photon energy, the linear
attenuation coefficient of water is 30 m-1. Then the CT number for this tissue in Hounsfield units is:
(Use definition: CT (H) = 1000*((ut-uwater)/uwater) =>( (33-30)/30) )*1000= +100
.
For the next questions assume the following general case:
One part of an electronic radiographic system (that we consider to be split into two main stages)
transforms a one dimensional (x-axis) varying X radiation energy fluence (x), into a signal
voltage V(x). This signal voltage representation is thence eventually transformed to a numerical
value N(x) in the next stage of the system.
The radiation fluence pattern is detected over a spatial length (in the x axis) of 20 cm.
Q34.
Write down the first six (6) NON zero spatial frequencies (corresponding to m = 1, 2, 3, 4, 5,6) in a
'constant + sine series' representation of the pattern are:
(Use f = 1/P, P =20cm, then need 1f, 2f, 3f, 4f, 5f,6f => 1/20, 1/10, 3/20, 1/5, 1/4, 3/10 cm-1 )
Q35.
Suppose the first 6 coefficients (corresponding to m = 0, 1, 2, 3, 4, 5) of the V(x) 'constant + sine series'
are divided by the first 6 coefficients of the corresponding (x) terms. Then if this first system stage
considered was ideal, write down any three possible ratio values
( All must be same ratio value for ideal property)
Q36.
The ratio of the first 6 coefficients ( m = 0, 1,2,3,4, 5) for V(x)/(x) gives values (increasing frequency
order) of :
100, 90, 95, 80, 81.
A similar set of ratios for N(x)/V(x) gives: 10, 9.5, 9, 8.5, 7.0, 6.8.
Then calculate the Normalized MTF value for the complete system for a spatial frequency of 0.15 cm-1
:
(Use MTF(f)sys = MTF(f)1*MTF(f)2, => MTF(0.3cm-1)sys = 0.8*0.85 = 0.68)