The energy of the beta particle and the type of contaminant.

Question

Want to see the full answer?

Answer 1

Question

Chapter 5, Problem 5.5P

To determine

The energy of the beta particle and the type of contaminant.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Iridium-192, a beta emitter, is used in medicine to treat certain cancers. The 192Ir is used as a wire and inserted near the tumor for a given amount of time and then removed. The 192Ir wire is coated with pure platinum, and is sold by diameter and amount of radiation emitted. One type of wire sold is 221 mm long, has a linear apparent activity of 30–130 MBq/cm, an active diameter (which is the iridium-192) of 0.1 mm, and an outer diameter (the platinum) of 0.3 mm. (a) How much 192Ir is in the wire if the apparent activity is 99 MBq/cm?Enter your answer in scientific notation: (b) If the wire is placed in the tumor for 60 hours, how much radiation is absorbed by the tumor?

CSDA range Determine the expected range of 2 MeV beta radiation in polyethylene, Aluminium and air. Use the ESTAR database on the NIST website to find the CSDA ranges (to 4 significant figures): R(px) for polyethylene = 9.375E-01 ✓g/cm² R(px) for Al= 1.224E+00 ✓ g/cm² R(px) for air- 1.094E+00 x g/cm² Looking at the 3 values you have found, the statement in the lab notes that the range is 'inversely proportional to the density p of the material and not dependent on its structure or any other properties' is a useful rough approximation because the px values are similar even if not exactly the same The values given are density times thickness (px). To find the actual range (x) we just divide by the density.

The linear absorption coefficient for 5-MeV gamma rays in lead is 78 m-1. Find the thickness of lead required to reduce by one- fourth the intensity of beam of such gamma rays. O A. 8.9 mm O B. 17.8 mm OC. 3.7 mm OD. 1.9 mm O E. 7.4 mm