Test 3 - NCRP 49 Terms

Question 1

Weekly design exposure rate (limit on exposure on person)

Answer 1

P, unit is in Roentgen

Question 2

Weekly workload X-ray

Answer 2

W, unit is in mA*min

Question 3

Weekly workload MV and gamma ray

Answer 3

W, unit is in R, measured at 1 meter from source

Question 4

Use factor

Answer 4

U. 1 for floor, 1/4 for walls, since most x-rays tubes point downward.

Question 5

Occupancy factor

Answer 5

T. 1 for commonly occupied spaces, 1/4 for medium, 1/16 for sparsely occupied.

Question 6

Normalized output

Answer 6

X_n. Exposure per current output. Unit given as R/mA at 1 meter from source.

Question 7

Exposure rate at 1 m from source of 1. useful beam, 2. leakage radiation, 3. scattered radiation

Answer 7

X’_u, X’_L, X’_s. (primes represent time derivative) Unit given as R/min at 1 meter from source.

Question 8

Quotient of exposure at 1 m and workload

Answer 8

K_UX. Unit given as R/mA-min at 1 meter from source. Also known as exposure per workload.

Question 9

Transmission factor for useful beam for 1. x-rays 2. gamma rays 3. leakage x-rays 4. leakage gamma 5. scattered x-ray 6. scattered gamma ray.

Answer 9

B_UX, B_UG, B_LX, B_LG, B_SX, B_SG. Transmission factor

Question 10

Barrier Thickness 1. primary wall 2. secondary wall

Answer 10

S_p, S_s. Primary wall is the actual irradiated wall for useful beams. Secondary wall is the wall irradiated by leakage and scattered radiations.

Question 11

Exposure in terms of X’_u and distance d in meters

Answer 11

X = X’_u*t/(d)^2. Since X’_u is defined as exposure rate at 1 meter, d in meter is suffice to obtain true exposure at point of interest.

Question 12

Solving for B using P, X’_u, t and d.

Answer 12

Since P is what we want to achieve, we can solve for B given the exposure limit, P. P = B * X, and we know X = X’_u * t / d^2. So, P = B * X’_u * t / d^2. Note P = 10 mR/week, as of NCRP 147.

Question 13

X’_u in terms of X_n, t, and workload W.

Answer 13

X’_u is exposure rate at 1 m. X_n is exposure at 1 m per workload W. Thus X_n * W / t = X’_u.

Question 14

Solving for B using P, X_n, and workload W.

Answer 14

Since P = B * exposure / d^2, and exposure = X’_n * W (since X’_n is R at 1 meter per workload 1 mA*min), P = B * X_n * W / d^2.

Question 15

Solving for K_UX given exposure limit P

Answer 15

Since K_UX is exposure / mAmin at 1 meter, K_UX = P * d^2 / W. Why? P is desired exposure at point of interest d meters away. P / W gives exposure per workload. Multiplying by d^2 will give R / mAmin AT 1 METER, or K_UX.

Question 16

Solving for K_UX with U and T, use factor and occupancy factor.

Answer 16

U and T will give leniency to allowed K_UX (specifically will raise K_UX). This means more workload is needed to reach P, so more leniency. Specifically, K_UX = P * d^2 / WUT.

Question 17

General equation relating B (transmission factor), X’_n (exposure per workload) to P (exposure limit), d, and WUT in the presence of shielding

Answer 17

K_UX = B*X’_n = P * d^2 / WUT
Note that B is missing if no shielding exists. Note that when one calculates P * d^2 / WUT, the value can be used on a table to search for right lead thickness for acceptable value.

Question 18

What was NCRP 49? Replaced by what NCRP? What was updated?

Answer 18

Structural shielding design for medical use of x-rays and gamma rays. NCRP 147. Dose limits for medical shielding designs were defined.

Question 19

NCRP 116 vs ICRP 60 on dose limits

Answer 19

To workers. Annual effective dose limit: 20mSv for both. Cumulative eff. dose: 10mSv x age VS 20mSv in 5 years. Equivalent dose: 150mSv/500mSv lens/rest for both.Dose to public. Eff. dose limit: 1mSv if continuous, 5mSv if infrequent VS 1mSv, higher if needed provided 5yr avg <1mSv/yr. Equivalent dose: 15mSv/50mSv lens/rest for both.

Question 20

NCRP 49 dose limit (old and defunct)

Answer 20

Derived from ICRP 26 limits by assuming 50 week = 1 working year. So, eff. dose limit for radiation worker is 100mR/wk ~ 1mSv/wk, for public is 10mR/wk ~ 0.1mSv/wk.

Question 21

NCRP 49 VS 147 x-ray machines occupational/public dose limit?

Answer 21

Occupational: 100mR/wk to 10mR/wk! (1mSv/wk to 0.1)Public: 10mR/wk to 2mR/wk! (0.1mSv/wk to 0.02)

Question 22

Rotating anode disk ~1930s

Answer 22

Molybdenum disk used. ~3000rpm, ~10000rpm for cineangiography. 6mm x 1.5mm = 9 mm^2 region bombarded with e-, with 1900mm^2 total area bombarded per rotation. Coolant oil in the covering of the tube absorbs infrared light.

Question 23

Aluminum filter restrictions

Answer 23

Under 50kVp: 0.5mm, 50~70kVp: 1.5mm, above 70: 2.5mm Aluminum.

Question 24

Beam hardening and effective/equivalent energy

Answer 24

Hardening = higher equivalent energy. Equivalent energy is monoenergetic beam that would go through same attenuation as the spectrum.

Question 25

PDD defined in radiation protection

Answer 25

PDD = D(x)/D(0) where D(0) is SURFACE dose. PDD is a function of SSD, effective energy, and area of irradiation.

Question 26

What does having larger field size do to PDD?

Answer 26

More field size = more scatter. Thus PDD is higher / depth profile is flatter.

Question 27

Rule of Thumb of HVL of the body for x-rays

Answer 27

HVL ~ 4cm. The body is around 20cm, or 5 HVL long.

Question 28

What is DAP?

Answer 28

Used for x-ray exposure. Dose area product (Gy * cm^2) good estimation of TOTAL energy delivered to body. Since "dose" is only E/mass, need to multiply by surface area to get better estimation of actual radiation energy delivered. Also, DAP is constant wherever the beam is measured, meaning DAP can be measured near the beam source to estimate dose to patients.

Question 29

NCRP 49 Shielding Terms

Answer 29

Distance, time, barrier, primary barrier, secondary barrier, workload (mA*min/wk), use factor (U), occupancy factor (T)

Question 30

Use Factor (U)

Answer 30

Fraction of the mA*s that is directed towards a particular barrier. Use factor for secondary wall is 1.

Question 31

Rule of Thumb for Dose for X-ray with certain kVp

Answer 31

Dose(1) / Dose(2) = [kVp(1) / kVp(2)]^2

Question 32

The filter used at Sands Building is:

Answer 32

#4 filter, 0.1mm Cu with 2.5mm Al.

Question 33

f-factor

Answer 33

Conversion directly from X to Dose to tissue. Dmed = f-factor * X. Formula: f-factor = [u(med)/u(air)] * 0.869 rad/R, where u is u(en), or mass absorption coefficient.If exposure is given in C/kg, use f-factor = [u(med)/u(air)] * 33.7 J/C

Question 34

Dose in terms of mass absorption coefficient?

Answer 34

D = [u(en)/p] * PSI, where PSI is energy fluence.

Question 35

Dose in soft tissue INCLUDED within the bone? (for example, osteocytes?)

Answer 35

f-factor greatly increased for lower energies. Also, larger the cavity, lower the f-factor. Lowest f-factor in the center of the cavity (electron build-up less).

Question 36

X-ray THERAPY Leakage Limit for above 500 kVp

Answer 36

0.1% of useful beam exposure RATE

Question 37

X-ray DIAGNOSTICS Leakage Limit for below 500 kVp

Answer 37

0.1 R / hr @ 1 meter (100 mR / hr)