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Treatment Planning Fill In The Blanks

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In each blank, try to type in the word that is missing. If you've typed in the correct word, the blank will turn green.

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When you are all done, you should look back over all your answers and review the ones in red. These ones in red are the ones which you needed help on.
Question: Therapy prescriptionAnswer: Communication tool between the radiation oncologist and the treatment and delivery team (medical dosimetrist and radiation therapist) and provides the information required to administer the appropriate radiation treatment (W/L, pg. 493).
Question: Components of RT Answer: Defines the treatment volume, intended tumor dose, number of treatments, dose per treatment, and frequency of treatment. Also stated are the type and energy of to be used, beam-shaping devices and any other appropriate factors (W/L, pg. 493).
Question: Dose (or dose)Answer: Refers to the energy deposited at a specific point in a medium. The dose is measured at a point (in a patient or phantom) and is commonly measured in Gray (Gy). (W/L, pg. 493).
Question: DepthAnswer: The distance beneath the skin where the prescribed dose is to be delivered (W/L, pg. 494)
Question: Answer: The measurement of the patient’s thickness from the point of beam entry to the of beam exit (W/L, pg. 494).
Question: SSDAnswer: The distance from the source of photons to the patient’s skin (W/L, pg. 494).
Question: SADAnswer: The from the source of photons to the machines isocenter (W/L, pg. 494).
Question: IsocenterAnswer: The intersection of the axis of of the gantry and the axis of rotation of the collimator for the treatment unit (W/L, pg. 494).
Question: SizeAnswer: Physical dimensions set on the collimators of the therapy unit that determine the size of the treatment field at a reference (W/L, pg. 494).
Question: Depth of Maximum (Dmax)Answer: The at which electronic equilibrium occurs for photon beams. Dmax is the point where the maximum absorbed dose occurs for single field photon beams and depends mainly on the energy of the beam (W/L, pg. 496).
Question: OutputAnswer: The dose rate of the ; it is the amount of radiation “exposure” produced by a treatment machine or source as specified at a reference field size and a specified reference distance. (W/L, pg. 496).
Question: FactorAnswer: Ratio of the dose rate of a given field size to the dose rate of the reference size (W/L, pg. 496).
Question: Gap Answer: Gap = (L1/2 × d/SSD1) + (L2/2 × d/SSD2) (W/L, pg. 516).
Question: DoseAnswer: Dose= (TD/PDD) × 100 (W/L, pg. 509).
Question: Dose (Dmax Dose)Answer: TD= (Given dose x PDD at of calculation)/100 (W/L, pg. 509).
Question: F-factorAnswer: (SSD1+ d)^2 / (SSD1+ Dmax )^2 ×(SSD2 )+ Dmax )^2/(SSD2+ d)^2 *New PDD = Old PDD x F-factor (W/L, pg. 508).
Question: Square LawAnswer: I1/I2 =(D2/D1 )^2 (W/L, pg. 497).
Question: FormulaAnswer: Square= (4(L ×W)) / (2(L +W)) (W/L, pg. 498).
Question: Square Correction Factor (ISCF) [for SSD Set-ups]Answer: ISCF = (Reference + Dmax)^2 / (Treatment distance + Dmax)^2 (W/L, pg. 508).
Question: ISCF (for Distance Set-ups)Answer: ISCF = (Reference source calibration distance)^2 / (Treatment SSD +Dmax)^2 *(Reference calibration distance = Reference distance + Dmax for the energy) (W/L, pg. 508).
Question: ISCF (for SAD Set-ups)Answer: (Reference distance / Source Calculation Point Dose [SCPD])^2 *Reference distance = 100; SCPD = The set up SSD + (W/L, pg. 508).
Question: Time/MU for SSD Set-upsAnswer: MU/Time=(Prescribed Dose)/ (RDR*ISCF*Sc*Sp*PDD/100*Other) (W/L, pg. 504)
Question: Unit Calculations for SAD (Isocentric) Set-ups (TAR)Answer: MU= (Prescribed Dose) / (RDR×ISCF×Sc×TAR×Other ) (W/L, pg. 511).
Question: Unit Calculations for SAD (Isocentric) Set-ups (TMR)Answer: MU= (Prescribed Dose) / (RDR×ISCF×Sc×Sp×TMR×Other ) (W/L, pg. 511).
Question: Unit Calculations for SAD (Isocentric) Set-ups (TPR)Answer: MU= (Prescribed Dose) / (RDR×ISCF×Sc×Sp×TPR×Other ) (W/L, pg. 511).
Question: Hinge Answer: HA = 180 - 2(wedge ) (RT Essentials, pg. 135)
Question: AngleAnswer: WA = 90 – (Hinge angle/2) (RT , pg. 135)
Question: Electron Beam Mean Answer: EO = C4R50; EO = Mean Energy; C4 = 2.4 MeV; R50 = Depth of 50% dose in cm (W/L, pg. 554).
Question: Practical Range (Er) in cm Beam in TissueAnswer: Er = MeV/2 (W/L, pg. 554).
Question: Electron 80% Isodose Answer: MeV/3 (W/L, pg. 555).
Question: Electron 90% lineAnswer: MeV/4 (W/L, pg. 555).
Question: Full Strength SourceAnswer: A = (0.66mg/cm) x (active of source in cm)
Question: Activity Half SourceAnswer: A = (0.33mg/cm) x (active length of in cm)
Question: Patient is to be with AP/PA fields (2:1). Total dose is 200 cGy. What is the dose to each field?Answer: 200 cGy/3 = 66.7 = 66.7 x 2 = 133.4 cGy. PA = 66.7 x 1 = 66.7 cGy.
Question: Patient is to be treated with total dose of 180 cGy in a four field arrangement AP, PA, RL and LL (2:1:1.5:1.5). What is the dose to each ?Answer: AP field: 30 x 2 = 60 cGy PA : 30 x 1 = 30 cGy RT Lat field: 30 x 1.5 = 45 cGy LT Lat field: 30 x 1.5 = 45 cGy
Question: Calculate the Gap: Field 1- Length = 17 cm, = 6 cm, Depth = 3 cm, SSD = 92 cm; Field 2- Length = 15 cm, Width = 12.5 cm, Depth = 3 cm, SSD = 91 cm.Answer: (17/2 x 3/92) + (15/2 x 3/91) (8.5 x .0326) + (7.5 x .0324) (.2771 + .0243) Gap= .52
Question: What is the size on a film if the collimator setting is 7 cm X 19 cm and the magnification factor is 1.33x?Answer: 9 X 25; 1.33 x 7 cm = 9.31 cm; 1.33 x 19 cm = 25.27
Question: A patient has a 3.5 cm wide at a depth of 4 cm. What will be the necessary field width at the skin surface to cover the tumor volume plus a 1 cm margin on each side using a linear accelerator and isocentric set-up? Answer: (5.5/100) = (x/96); 100x = 528; x = 5.28.
Question: The dose rate on a accelerator is 102.4 cGy/Min at 100 cm. What is the dose rate at 85.5 cm?Answer: (102.4/x) = (85.5/100)^2; x = 140 cGy/Min.
Question: The optimum hinge angle for a 60 degree wedge be:Answer: 180 - 2(60) = 60 .
Question: Calculate the Gap: Field 1- Length = 10 cm, Width = 10 cm, Depth = 5 cm, SSD = 100 cm; Field 2- Length = 15 cm, = 10 cm, Depth = 4 cm, SSD = 100 cm.Answer: (10/2 x 5/100) + (15/2 x 4/100); Gap= .55cm.
Question: Calculate the square for a field size of 10 cm X 15 cm.Answer: (4(10x15) / (2(10+15) = 12.
Question: What are the 80% and 90% lines for a patient treated with a 16 MeV electron beam?Answer: 80% = 16/3 = 5.3 cm; 90% = 16/4 = 4 cm.
Question: A patient is treated at 100 cm SSD with 6 MV photons. Collimator setting is 15 cm x 15 cm. There is no blocking. 300 cGy per is to be delivered at Dmax. What is the dose delivered at a depth of 5 cm?Answer: (300 x 87.9)/100 = 263.7 cGy. (See 24-6 W/L pg. 519)
Question: A patient is prescribed a dose of 180 cGy at a depth of 10 cm with 10 MV photons at 100 cm SSD. The PDD is 60%. the dose to the depth of maximum dose.Answer: 180/0.60 = 300 cGy.
Question: What is the between the central rays of the two beams when using a wedged pair?Answer: Angle.
Question: A 45 degree wedge is inserted into a field to modify the isodose curve. The toe section will allow (greater or lesser) in part of the beam.Answer: .
Question: What type of corrections must be made to for bone, tissue and muscle when doing calculation? Answer: Heterogeneity .
Question: Palpable tumor; visible areas of disease.Answer: GTV.
Question: volume which allows for patient motion and set up uncertainties.Answer: PTV.
Question: The area enclosed by the surface selected.Answer: volume.
Question: Contains a margin for subclinical extensions of the .Answer: CTV.
Question: The point A used when calculating dose for cervical and uterine treatments is located:Answer: 2 cm superior and 2 cm lateral to the of the cervical os.
Question: If setting a 10 x 10 field size using an technique, the field size on the patient’s skin would be?Answer: .
Question: The most widely used substitutes is:Answer: -137 (W/L, pg. 303).
Question: Which radium substitute would be best suited to implants of the breast and tongue?Answer: -192 (W/L, pg. 305).
Question: What is the half-life of -226?Answer: 1,622 (W/L, pg. 303).
Question: What is the half-life of -60?Answer: 5.27 (W/L, pg. 303).
Question: What is the half-life of -137?Answer: 30.0 years (W/L, pg. 303).
Question: What is the half-life of -192?Answer: 73.83 days (W/L, pg. 303).
Question: What is the half-life of -125?Answer: 59.4 days (W/L, pg. 303).
Question: What is the half-life of -103?Answer: 16.99 days (W/L, pg. 303).
Question: What is the half-life of gold-198?Answer: 2.7 days (W/L, pg. 303).
Question: What is the half-life of -222?Answer: 3.82 days (W/L, pg.
Question: A patient is being treated at 115 SSD on a 100 cm SAD . The collimators have been set to 35 cm X 40 cm. What is the field size on the patient’s skin?Answer: 35/100 = x/115; 40/100 = y/115; 100x = 4025; 100y = 4600; x = 40.25 y =
Question: What is the of bolus?Answer: Fills in deficits to have a more homogenous dose distribution. Shifts dose lines and brings Dmax closer to the skin surface when skin sparing is not desirable (Mosby’s RT Guide, pg. 102).
Question: For non-isocentric treatments, _______ is the factor of choice to demonstrate central axis dose at a given depth.Answer: %DD or PDD (W/L, pg. 509).
Question: When looking up the PDD or TMR for a given depth and field size, _______ should be used when there are blocks or MLC.Answer: Square (W/L, pg. 504).
Question: HDR isotopes deliver at a dose rate =_______cGy/min Answer: 20 cGy/min. (Mosby’s RT Guide, pg. 108).
Question: LDR isotopes deliver at a dose rate =_______cGy/min Answer: 0.5 to 2.0 cGy/min. (Mosby’s RT Guide, pg. 108).
Question: The wedge angle is determined by the tilt of the isodose lines at_______.Answer: The 50% isodose line in low energy beams like 60 or the isodose line at a depth of 10 cm for higher energy beams used in modern linear accelerators (Mosby’s RT Study Guide, pg. 101).
Question: What is the Dmax for a 1.25 MV beam?Answer: 0.5 cm (RT , pg. 140).
Question: What is the Dmax for a 4 MV beam?Answer: 1.0 cm (RT , pg. 140).
Question: What is the Dmax for a 6 MV beam?Answer: 1.5 cm (RT , pg. 140).
Question: What is the Dmax for a 10 MV beam?Answer: 2.5 cm (RT , pg. 140).
Question: What is the Dmax for a 18 MV beam?Answer: 3.5 cm (RT , pg. 140).
Question: What is the Dmax for a 24 MV beam?Answer: 4.0 cm (RT , pg. 140).
Question: Treatment planning systems often combine CT images with images from other modalities. What is this ?Answer: fusion or image registration (W/L, pg. 542).
Question: What is the practical in tissue for a 10 MeV electron beam?Answer: Er =MeV/2; Er = 10 MeV/2; Er = 5
Question: A dose of 200 cGy/fraction is to be to a depth of 10 cm using an AP:PA treatment arrangement. The fields are weighted 3:2 AP:PA. What is the dose per field?Answer: 200 cGy/5 = 40 cGy; AP = 40 x 3 = 120 cGy; PA field = 40 x 2 = 80 cGy.
 
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