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Skin Radiation therapy > Tx planning > Flashcards

Tx planning Flashcards

1
Q

1 mm of lead stops how much Mev

A

every mm of lead stops 2MEv of photons

therefore if you have a 6Mev beam you will need 3 mm of lead to stop the beam

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2
Q

how to calculate activity

A

A= A0 e ^-uxt

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3
Q

whqat does u in an activity/ half life problem represent

A

it is .693/ half life time

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4
Q

practical range for electrons

A

MEV /2

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5
Q

what % lin do we prescribe to in electron

A

90%

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6
Q

Mev coverage of 90% = Depth x ____

A

4

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7
Q

mev coverage of 80% = depth x ___

A

3

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8
Q

mean nrg of electron beam in mev at surface =

A

depth in cm at 50% iso line x 2.4

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9
Q

what is the therapeutic ratio

A

graph between the response rate( y asix) and dose (x axis) can determine the optimal therapeutic ratio
the response curve of normal tissue vs tumour is plotted on graph
-the therapeutic ratio is the shoulder of the curve btwn the dose and response

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10
Q

what isx the optimal therapeutic ratio

A

where there is the greatest distance btwn tumour and normal tissue

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11
Q

what does v20 mean

A

it is the v (volume) of a certain organ receiving a certain dose v20 is the volume 20Gy- and is associated with a percent
therefore if lung v20 <30%
it means <30% of th lung should get 20Gy

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12
Q

when photons reach the pt they interact mostly with ____.

A

electrons

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13
Q

radiation that is scattered back towards the pt is called____.

A

backscatter

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14
Q

dmax 6 mv

A

1.5cm

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15
Q

dmax 10 mv

A

2.5 cm

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16
Q

dmax 18 mv

A

3.5 cm

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17
Q

dmax for kv beams

A

at the surface of the skin

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18
Q

what can influence dmax

A

fs and distance

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19
Q

in high energy MV beams the dmax helps with what issue?

A

skin sparing- the dose is beneath the surface of the skin

ex: 6MV the dmax is 1.5 cm below surfac of skin

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20
Q

in rectum plan wedges sre plaed with thick ends ____ because_____

A

thick ends are post (bottom) because it is attempting to even out dose distribution not being used as missing tissue compensator as is seen in the breast

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21
Q

ISL is a relationship between what

A

the distance and intensity

they are inversely proportional

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22
Q

what is the effect on the intensity/ dose when the distamce is doubled

A

ISL

therefore the new intensity will be 1/4

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23
Q

which is LARGER Effective FS or collimated FS

A

EFS is smaller than collimated FS

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24
Q

+ nrg PDD_

A

+

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25
Q

FS + PDD _

A

+

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26
Q

DISTANCE + PDD_

A

+

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27
Q

What is the equation for gap calc

A

(L1/2) x (d/SSD) + (L2/2)x(d/SSD)

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28
Q

How to calculate hinge angle

A

180 - (given < x2)

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29
Q 
whats maynord factor for pt who was originally tx at 100ssd to d=5cm w 10mv photon beam but now needs to be tx at 110cm ssd
A. .990
B: .993
C: 1.003
D: 1.010
A

C

(SSD1 +D/ SSD1+DMAX )^2 X (SSD2 +DMAX/SSD2+D) ^2

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30
Q 
Calc gap req for adjacent fields Field a length is 12 field b length is 20 tx at 100 ssd with depth of 5 cm
A: .5cm
b: .8cm
c: 1.3cm 
d: .1.5cm
A

B

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31
Q 
how many MU req for lumpectomy boost electron prescribed to 80% iso line the Cfs for tx field using 10x10cm cone with min blocking is 1.05. the dose/ field is 200cGy
A: 211MU
B: 238MU
C: 190 MU
D: 214MU
A

B

Dose / cf x PDD x ccal

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32
Q

the effect on the PDD is greater with _ wedge angle and _ beam energy

A

+ wedge abgle

- beam nrg

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33
Q

the term used for when collimation is used to create wedging in the field is _____

A

dynamic wedge

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34
Q

____ this type of wedging uses same angles as physical wedges`

A

dynamic wedge

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35
Q

this type of wedging only uses wedge angle of 60 degrees

A

virtual wedge

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36
Q

_____ this type of wedging uses a 60 degree wedge and fields are partially wedged if a wedge smaller than 60 dgrees is needed

A

virtual wedge

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37
Q

calculate wedge angl given hinge anglev

A

wedge angl = 90- hinge angl/ 2

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38
Q

a photon has _ charge and _ mass

A

none

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39
Q

an electron has _ charge and _ mass

A
  • charge and the mass of 1/2000 of the mass of proton
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40
Q

the likelihood of interacting with atoms is greater in a photon or electron?

A

electron

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41
Q

mass stopping power is affectd by what 2 things?

A

the energy of the electron beam and the z (atomic #)of irradiatedd material

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42
Q

____ materials have high electron density

A

low z material

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43
Q

what 3 things must occur in order to change from photons to electrons in the LINAC

A

1 remove target and flattening filter
2 decrease the electron gun curent
3 add the scattering foil nd scanning elevtron beam

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44
Q

a narrow eldctron beam created by th linear accelerator is called_____

A

pencil beam

45
Q

why is the electron gun current - when switching from photons to electrons

A

the beam of current through the electron window is 1/1000 of the beam current at he xray target for xray therapy mode
without he reduction of the curent at the eldctron gun window dose rates of 1000s of centigray could result

46
Q

what is used in order to transform a pencil beam into a usable beam for treatment

A

a scattering foil is used to make pencil beams of electrons usable

47
Q

what does the first scattering foil do? the second?

A

first widens the beam and the second improves flatness

48
Q

what is required for the scattering foil to be effective

A

be made of high atomic (z) material

49
Q

when are scanning eldctronbeams most useuful

A

at energies > 25MEV as at this dose usingb scattering foil at this dose requires v thick foil which results in mechanical difficulties and can cause probs with electron contamination

50
Q

cons of scanning electron beams

A
  • they can have xray contamination

- failures of the mechnisim can cause devestating results to the pt

51
Q

cons of scattering foils

A

bremstralung contamination

52
Q

contraindications in treatment with orthovoltage and superficial x rays

A

you can not use them when there is bone directly underneath

53
Q

what is the consant used`in detrmining mean dose

A

2.33Mev

54
Q

equation for mean energy

A

E0= depth of 50% dose in cm x c4 (2,33MEV)

55
Q

what is the mean energy when 50% of dose is delivered in 2.5cm of tissue?

A

E0= 2.5 x 2.33MEV= 5.85mev

56
Q

is a 6MeV beam sufficent to treat 3.5cm of tissue? why or why not?

A

think of practical range
e= meV/2
6/2 = 3cm therefore 6MEV is only able to treat to 3cm of tissue therefore 6MEV is not sufficient in this case

57
Q

what is the depth at iso for an 18mev beam given its prescribed to 90%

A

18mEv/4 =4.5cm

58
Q

if given a depth of 5cm prescribed to the 80% isodose line what is the enrgy required

A

5cm=Mev/3
5cmx 3=mev
15mev is required in this case

59
Q

in electron beams as energy if the beam + surface dose_

in photon beams as energy + the surface dose_

A

electron the surface dose +

in photon the surface dose -

60
Q 
A patient has a lesion on thir leg that the RO wants to tx with electrons with a beam energy  of 9mEv, what FS is appropriate?
A) 3cm
B) 4cm 
C)2cm 
D) 6cm
A

D 6cm
this is bcause the FS must be larger than the size of the practical range
N.B. practical range is energy/2 therefore = 4.5cm
since it mucst be largr than 4.5cm the only appropriate answer is D

61
Q

what is never done in terms of using bolus with electron fields

A

partial bolusing is never done, a decrease in dose is possible when doing bolus with electrons

62
Q

in edge effect unbolused areas have _ dose and bolused areas have_ dose, how much can this dose fluctuate by

A

in unbloused areas the dose can + and in bloused areas the dose can - this dose can fluctuate by 20-30%

63
Q

the 3 uses of bolus in electrons

A

think IRS
I- incfease dose to surface
R-remve surface irregularities
s-shape iso contours at depth

64
Q

use this to calculate the thickness of lead for sheilding electrons

A

the MEV of the beam / 2= thickness of lead +1mm

65
Q

how is the thickness of cerrobend calculated for electrons

A

take the thickness in lead x1.2

66
Q 
a 12MEV electron beam is being used and shielding with both cerrobend and lead is required what is the thickness of each?
A: 8mm lead and 7.4mm cerrobend
B: 7mm of lead and 8.4 mm of cerrobend
c: 10mm of lead and 12mm cerrobend
d: 6mm of lead and 7.22 mm of cerrobend
A

B
n.b. the thickness of lead is =mev/2+1mm
thickness of cerrobend is thickness of lead x1.2

67
Q

in order to have appropriate shielding for electrons what is required?

A

th dose needs to be shielded to 5% of the total dose

68
Q

what is the danger in internal shielding in electrons

A

dose to the tissues directly in front of shield may be + by 30-70%b/c of the electron back scatter from the shield

69
Q

what should internal eye sheilds be made of

A

low z material ex: tungsten

70
Q

in photons how much of the dose results from scatter?

in electrons how mych of the dose results from scater?

A

in photons 10-30% results from scatter

in electrons almost ALL absorbed dose results from scatter

71
Q

what happens to the dose when the diameter of the electron beam is < practical range

A

the dose is -

72
Q

what is the effect of dose on the bone in electrons

A

dose to tissues behind sm bones are - as might be expected b/c of shielding effect of the bone.
dose to tissues lateral to the bones is + bc lateral scattering of electron beam

73
Q

what is effect of the dose on the air cavity in electronsq

A

doe in air cavity is + as a result of - scattering of electrons as they psss through air cavity

74
Q

under/ ovrdose in % of electrons in air caviy and bone

A

20% underdose behind bone

15-35% overdose bhind cavity

75
Q

if partial bolusing in electron field is required what modification is required for the bolus?

A

the edges of the bolus should be tapered

76
Q

minimum width of bolus on a scar boost for electrons?

A

should be at least 2cm to ensure buildup of dose

77
Q

maximum FS for electron beam

A

25cm x 25cm

78
Q

what is an application of electron and photon junctioning of fields

A

iM chain in bRCA is treated with electrons while also getting a photon tangent fields

79
Q

IN JUNCTIONING A PHOTON WITH AN ELECTRON FIELD THE ELECTRON FELD SHOULD BE PLACED WHERE IN RELATION TO PHOTON FIELD

A

the electron field should be angled away from the photon field in order to match the divergence of the photon beam

80
Q

_ % of target receives the prescription dose

A

95%

81
Q

V20 lung typically

A

<35%

82
Q

how large is the PTV typically

A

CTV+.5cm

83
Q

if 10 MV photons were used to tx tumour to 100cGy then compared to 4Mv beam tx plan which of the follwoihng would be true if all other factors stay the same?
A. rectum will get + dose
B. bladder will get lower dose
C. rectum will get lower dose
D. bladder will get lower dose
E. no difference in rectum and bladder dose

A

a and d
this is because with higher energy more dose is deposited in deeper levels of tissue and more tissue is spared near the surface

84
Q 
GTV is at 10 cm of depth dose of 100cGy is given to GTV using 4MV beam bladder is at 7cm depth, ASSUMING SSD tech, which of the following is true?
A. rectum will get >100cGy
B. bladder will get >100 CGY 
c. rectum will get <100cGy 
d. bladder will get <100cGy
e. all of the above
A

b,c
using the SSD technique less dose is absorbed as depth + therefore tissues before the target will get more dose and tissues beyond the target will get - dose

85
Q

patient delivered dose of 150cGy d=13cm with 10Mv @100SSD with PDD= 60% what is dmax dose

A

dmax dose= total dose/ PDD
150cGy / 60%
dmax dose= 250cGy

86
Q

what type of brachy is the follwoing?

  • pros seed implant
  • an APBI implant ex: mammosite
  • Au 198 seeds in the tongue
A
  • Interstitial
  • intraluminal
  • interstitial
87
Q

how do you calculate exposure when given distance for a particular isotope

A

xposure= (activity x decay constant) / distance 2

88
Q

what is exposure for a 500mCi source of ir-192 with 1/2 life of 5.27 years at 5cm distance

A 
exposure= 500 x 5.27 / 5 ^2
activity = 105.4
89
Q

a primary br tumour located in rt temporal lobe will likely be tx with?
a: two paralel opposed = wt photn beams
b. wedged pair of RPO and ant field
C. wedged field consiting of vertex and rt lat field
d. 4 field box with wedges on lat

A

c
rt temporal tumour is a challenge due to proximity of the lens, thrfore vertex field is used as it is able to spare lens more than an ant field will

90
Q

to avoid exceeding the dose lims of the kidneys, the _____ technique is often employed when tx the pancreas?

a: 3 fld
b: 4 fld
c. post oblique
d. wedged field

A

a 3 fld tech with RT/ LAT/ AP r employed when tx the pancreas to - dose to post locared kidneys

91
Q

equation for couch kick

A

couch kick in degrees = tan -1 A(length tx field) /2x SSD

92
Q

What is the degrees of a couch kick for a pt with a 10cm x 12cm tx field treated at 97 cm SSD

A

Couch kick = - tan-1 (12/ 2x97)

couch kick = 3.4 degrees

93
Q 
an en face electron boost to lumpectomy bed is accidentally tx at 105cm SSD instead of 100 SSD the error in dose delivered is ?
A. 5% underdose 
B. 5% overdose
C. 10% underdose
D. 10% overdose
A

C

ISL (d1/d2) ^2
100/105 ^2
=.907
therefore underdose

94
Q

calculate the extended SSD for a 6 ft 7 inch ptg getting inverted Y treatment, initially the tx field is 100 cm SSD and a tx field of 40cm, however due to the pts height he requires 45cm tx field, what is the new extended field SSD

A

SSD new= SSD old x (d new/d old)
SSD new = 100 (45/40)
SSD new = 112.5 cm

95
Q

which is generally largr? FS or collimated FS ?

A

Fs largr than coll FS

96
Q

calculate the collimated FS for a 8x42 cm FS using xtended SSD of 115cm

A

Fs = 8cm x 100cmSSD/115cm SSD = 6.9CM
fS 42cm x 100cm SSD /115cm SSD = 36CM
Therefore given the FS the collimated FS = 36 cm x 6.9cm

97
Q

Calculate the field size given collimated Fs of 12cm x30 cm for extended SSD of 130cm

A

FS = 12x 130cm/100cm = 15.6cm
Fs= 30cm x130cm/ 100cm = 39cm
therefore given the collimated FS the FS is 15.6cm x 39cm

98
Q

time calculation for orthovoltage calc

A

time = dose/ dose rate x BSF FS / BSF cone

99
Q

What is the exposure of a 25mCi seed of Au198 after 3 hrs at the distance of 10cm
n.b. the activity of Au 198 is 2.38

A

X= Activity x decay constant of Au/ d ^2 x time
x = 25mCi x 2.38 / 10^2 x 3 hours
X= 59.5/ 100 x 3hours
X =1,785 R

100
Q

A Cs source that initially has the 1000mCi activity what is its activity after 2 years?
n.b. Cs 1/2 life is 30 years

A 
A= A0 . e ^-u x t
A= 1000 e ^-(.693/30) 2
A= 955mCi
101
Q

what is the mg equivalent in radium of a 104mCi of Au198

n.b. th exposure rat constant of Au =2.38

A

mg ra = A x (exposure rate Au/ exposure rate Ra)
Ra- constant is 8.25
threfore = 104 x (2.38/8.25)
mg ra = of Au = 30mg Ra eq

102
Q

an implant requires 10mg Ra = radiactive material using Cs137 what amount of Cs is required?
nb the exposure rate constant of Cs is 3.26

A

constant Ra/ constant cs = 1 mg ra equivalent x 10 mg Ra
= 8.25/3.26 = 1mg Ra = x10 mg Ra
= 25.3 mCi

103
Q 
Question 1: A new batch of Iridium 192 has been accepted and the dosimetrist is about to check the calibration. The label states that the activity of the source was .351 mCi/seed 10 days ago. What activity can the dosimetrist expect today (HL= 74.2 days).
A) .002 mCi
B) 9.97 mCi
C) .320 mCi 
 D) .007 mCi
A

C
a=a0 e ^u x t
a= 351 e ^ .693/74.2 x 10
a= .320

104
Q

whats the tx timed for an ortho tx ?

Pt receives 4000cGy/10 fx with 5cm Fs and 10cm cone. SSD output is 18.83 cGy/s with 1 s shttuer corrn

A 
time= dose/fx / output - shutter corrn
ttin = 4000/10 / 18.83 - 1s
time = 20s
105
Q

whats pneumbra for electrons

A

1 cm

106
Q

whats the coefficent equivalent thickness for bone? lung?

A 
bone = 1.65
lung = .5
107
Q

calculate the Mu for an electron with an applicator size= 8x16 cm @ 95cm SSD using 15mEV electrons to deliver 300cGy a 90% isodose line
n.b. the ROF= .971

A

mu =dose/ PDDx ROF
PDD =.9 BECAUSE WE ARE PRESCRIBING TO 90% ISO LINE
MU= 300CgY/ .9 x .971
Mu = 343

108
Q

what is the MU for2.5 cm thick tumour 1.5cm below surface RO wants 1000/5 tx at 95% 10x 10 cm cone used with 9x6cm insert what energy is used? calc MU
ROF = .986

A 
mu= dose/ ROF x PDD x CF
MU = 1000/5 / .986 x .95 .1cGy/Mu
MU= 200/ .9367
MU= 213
109
Q

what is the dose point a in the following problem?
4000cGy is prescribed to pt A using 6MV beam nrg in a path of 12cm of lung the physical depth is 15cm and radiological depth is 10cm and the u = .05cm -1

A 
first find correction factor
= e ^ u (d-d1)
e ^ .05 (15-10)
e^ .25
= 1.28
apply corrn factor to dose = 4000 x 1.28 =5136cGy

Skin Radiation therapy (40 decks)

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