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ABR Physics Part 2 Exam Questions With Complete Solutions

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ABR Physics Part 2 Exam Questions With Complete Solutions

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ABR Physics Part 2 Exam Questions With Complete
Solutions

%dd(10) Correct Answers Percent depth dose at 10 cm.
Determined at time of commissioning. Scan PDD from bottom
to top of tank and shift 0.6*rcav upstream. For energies greater
than 10 MV, add a 1mm lead foil near 50 cm from the target to
reduce electron contamination or use equation in TG 51.

Accoustic Impedance Correct Answers Z=density x speed of
sound in medium

Alara levels Correct Answers ALARA level 1: 10% the
allowed limit per quarter
ALARA level 2: 30% the allowed limit per quarter

Amount of lead needed to shield electron beam to less than 5%
Correct Answers divide the practical range by 10 gives amount
of lead in cm

Approximate attenuation rates Correct Answers 6X = 3% per
cm
10X = 2.5% per cm
18X = 2% per cm

Approximate density of materials Correct Answers Air -
0.0012 g/cm3
Lung - 0.33 g/cm3
Tissue/water - 1.0 g/cm3
Bone - 1.8 g/cm3
Concrete - 2.35 g/cm3 (can vary significantly with composition)

,Steel - 7.85 g/cm3
Cerrobend - 9.4 g/cm3
Lead - 11.34 g/cm3
Tungsten = 19.25 g/cm3

Approximate shielding Correct Answers X-ray - 1 mm lead
CT - 2 mm lead
PET - 1 cm lead
Cs-137 - 3 cm lead
HDR - 50 cm concrete
Co-60 - 70 cm concrete
LINAC/Cyberknife - 170cm concrete

Approximate Z of materials Correct Answers Air - 7.6
Tissue - 7.4
Bone - 14.0
Prostheses - 20-25

Au 198 Correct Answers energy = 420 keV
Γ= 2.38 (R cm2mCi-1hr-1)
Half-life = 2.7 days
HVL lead = 0.3 cm

Background dose Correct Answers 2 mSv/year

Backscattered radiation has a maximum energy Correct
Answers 0.255 MeV.

Barn Correct Answers 1 barn = 1 10^-24 cm^2

BED Correct Answers Biologically effective dose

, BED=nd(1+d/(alpha/beta))

n - the number of fractions delivered.
d - is the dose per fraction (Gy).
alpha/beta- is the alpha-beta ratio for the tissue of interest
(tumor or healthy cells).

BEIR V mortality risks Correct Answers from an acute
exposure is 8% per Sv and for chronic exposures, it is estimated
at 4% per Sv

Bracy medical events Correct Answers Wrong patient, site or
nuclide
Total dose differs by 20%
Single fraction differs by 50%

Co 60 Correct Answers Average energy = 1250 keV
Γ= 13.07 (R cm2mCi-1hr-1)
Half-life = 5.3 years
HVL lead = 1 cm

Compton scatter proportionality Correct Answers Z
independent, proportional to electron-density.

Concrete shielding TVLs Correct Answers 6X - 37 cm
18X - 45 cm

Cone factor Correct Answers The cone factor is used to correct
the output difference between the standard cone size (e.g. 10x10
cm2) and the utilized cone size.
All cone factors are normalized to a 10x10 cm2 cone.

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