RPT 113EXAM 2Page 1

NAME: ______June 26, 2008

  1. You’re getting ready to perform a loose surface (smearable) contamination survey in a room that has a few detectable radiation sources. Is your first consideration the radiation exposure you may accumulate during the survey or the possibility of becoming contaminated? Why (in 15 words or less)? (10 pts)
  1. The pancake frisker is a great beta radiation detector. We used it to determine that a 210Po source emitted no beta radiation and we used a “Stretch Scope” to determine that the same source emitted no gamma radiation. Could the pancake frisker detect any alpha particles at all?
  1. Yes / No (5 pts)
  2. And if so, under what conditions? (5 pts)
  1. What is the area that is “smeared” in an industry standard loose surface contamination survey? Units are important. (5 pts)
  1. If a smear is taken that has greater than 50,000 cpm and therefore exceeds the common frisker scaler, what kind of instrument would be used to determine the contamination level? (5 pts)
  2. Scintillation detector
  3. Ion chamber
  4. Extendable GM detector
  5. Proportional detector
  1. Back to the 210Po source we used. That radionuclide has a half-life of 138 days and was assayed when we bought it at 0.1 μCi (microcuries) of activity. If we only have 0.0177 μCi left now, when was the assay date (in other words, when did the source have the full 0.1 μCi)? How many days decay is a good enough answer. (11 pts)
  1. John Doe has to conduct a radiation and contamination survey of the research reactor lab. Match the best instrument type with the surveys he will conduct. (12 pts)
  1. Ion chambera.Smear counting
  2. GM Extender (“Stretch Scope”)b.Fixed contamination
  3. Pancake friskerc.Radiation areas
  4. Gas-flow proportional counterd.Find hot sources
  1. How does a Gas-Flow Proportional counter differentiate between alpha radiation and beta radiation? (5 pts)

a. Beta radiation produces a greater pulse because of its greater penetrating ability

b. Alpha radiation produces a greater pulse because of its higher energy

c. Beta radiation produces a greater pulse because it carries a greater percentage of its decay energy than alpha radiation does.

d. It does not differentiate between alpha and beta radiation

  1. For the Radiological Environmental Monitoring Program (REMP), the Technician is required to collect a number of air samples. Those samples are collected weekly using air samplers, pumps that draw a measured volume of air through a filter paper. If the first analysis done on the filter paper is for “gross alpha/beta,” what type of laboratory detector/counter would the Tech use? (5 pts)
  1. Gamma radiation interacts in three different modes depending on the photon energy. Match the following modes with the energy levels. (6 pts)
  2. Pair production1. low energy
  3. Photoelectric effect2. medium energy
  4. Compton scattering3. high energy
  1. In the above three modes of gamma interaction, one of those modes requires a very defined minimum energy to take place. What is the mode of interaction and what is the minimum energy? (10 pts)
  1. For all of the typically used survey instruments we’ve discussed, what are three of the actions you need to take before you leave the instrument shop and head out to conduct a survey? (12 pts)

1.

2.

3.

  1. Gas filled detectors have three regions that are used for the operation of three different types of detectors. Name the three regions in the graph below, the regions of interest are identified as A, B, and C. (9 pts)

A.

B.

C.

SOME IMPORTANT EQUATIONS AND CONSTANTS

Avogadro's Number = 6.02 x 1023 atoms/mole

1 Ci = 3.7 x 1010 Bq (disintegrations/sec)

1 Ci = 3.7 x 104 Bq

1 Sv = 100 rem1 Sv = 1000 mSv

1 Gy = 100 rad1 mGy = 1000 mGy

1' (foot) = 30.48 cm

Decay Equation:A = Ao e -t

Specific Activity:SA = Nwhere N is the number of atoms

recall N = atoms per mole/grams per mole;

Avogadro’s Number is the atoms or molecules per mole

 must be in sec-1

Gamma Attenuation:I = Io e -x

with  as the linear attenuation coefficient, x is thickness of the shield

Inverse Square Law: Exposure rate from a point source falls off with one over the distance squared.

( ERx ) / ( ERy ) = (Dy)2 / (Dx)2

Where ER is exposure rate at either x or y distance, and D is the distance, either x or y.

6CE “rule” says dose rate at 1 ft = 6 x curies x emitted gamma energy (in MeV)