Question

Find LET1 keV for 10-MeV protons in (a) soft tissue, (b) bone of density 1.93 g...

Find LET1 keV for 10-MeV protons in (a) soft tissue, (b) bone of density 1.93 g cm–3

Homework Answers

Answer #1

The term “linear energy transfer (LET)” is used to indicate the average amount of energy that is lost per unit path-length as a charged particle travels through a given material. The LET for electrons is traditionally expressed in units of MeV/cm, or, when divided by the mass density, in units of MeV-cm^2/g. Values of the LET for electrons have been tabulated for many materials and for a wide range of energies of the incident electrons.

The average amount of energy deposited in a thin sample, per electron, can be estimated by multiplying the LET by the sample thickness, t. Similarly, the total energy deposited per gram of a specimen, following an exposure of N electrons/area, is

E=LET⋅Nρ

where ρ is the mass density of the specimen material.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Soft tissue has a linear attenuation coefficient in the range of ? = 0.35 cm?1 at...
Soft tissue has a linear attenuation coefficient in the range of ? = 0.35 cm?1 at 30 keV and ? = 0.16 cm?1 at 100 keV. For this problem we use ? = 0.21 cm?1, which applies at around 50 keV incident X-ray energy. (a) What fraction of X-ray photons at 50 keV are passing through a person's body? Hint: the person's body thickness is about 19 cm. Use a soft tissue approximation, that is, neglecting bones. N N0 =...
Compute the contrast between soft tissue (approximated as having the same attenuation coefficient as water) and...
Compute the contrast between soft tissue (approximated as having the same attenuation coefficient as water) and a 2 mm diameter artery filled with an iodine contrast medium at a density 400 milligram/cm 3 . Use photon energies of 40 Kev just above the iodine K absorption edge. Use values of: µ mwater = 0.36 cm 2 /g and µ mIodine = 22.38 cm 2 /g for 40 keV x-rays with (a) no scattering and (b) 50% scattering fraction. Scattering fraction...
If the half-value layer of cortical bone for positron generated gammas (511 keV) and average x-ray...
If the half-value layer of cortical bone for positron generated gammas (511 keV) and average x-ray CT energies (80 keV) are 38.8 and 17.3 mm, respectively, what is the ratio of tissue thicknesses needed in PET and CT to produce equal photon flux in both cases? What would the fractional attenuation (I/I0) be for 3 cm of cortical bone?
Find the momentum of a 13.6-MeV gamma ray; a 28-keV X ray; a 4.0-μm infrared photon;...
Find the momentum of a 13.6-MeV gamma ray; a 28-keV X ray; a 4.0-μm infrared photon; a 213-MHz radio-wave photon. Express the momentum in kg · m/s and eV/c. (a) 13.6-MeV gamma ray (b) a 28-KeV X ray (c) a 4.0-μm infrared photon (d) a 213-MHz radio-wave photon
. Assume that a randomly selected subject is given a bone density test. Bone density test...
. Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score or probability corresponding to the given information. Round results to two decimal places. (a) Find the bone density scores that can be used as cutoff values separating the lowest 3% and the highest 3%. The score...
How much energy is released per minute by the 14N7(n, p)14C6 reaction in a 10-g sample...
How much energy is released per minute by the 14N7(n, p)14C6 reaction in a 10-g sample of soft tissue bombarded by 2 X 1010 thermal neutrons cm-2 s-1? Nitrogen atoms constitute 3% of the mass of soft tissue.
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x...
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. Cola Consumed Bone Mineral Density (g) 1 0.8777 2 0.8925 3 0.8898 4 0.8769 5 0.8999 6 0.8634 7 0.8762 8 0.8888 9 0.8552 10 0.8546 11 0.8762 Based on your results, If the Colas Consumed increases by 1,...
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x...
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. Cola Consumed Bone Mineral Density (g) 1 0.8777 2 0.8925 3 0.8898 4 0.8769 5 0.8999 6 0.8634 7 0.8762 8 0.8888 9 0.8552 10 0.8546 11 0.8762 Based on the data in the table, is there a significant...
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x...
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. Cola Consumed Bone Mineral Density (g) 1 0.8777 2 0.8925 3 0.8898 4 0.8769 5 0.8999 6 0.8634 7 0.8762 8 0.8888 9 0.8552 10 0.8546 11 0.8762 Based on your results, If the Colas Consumed increases by 1,...
A positron emerges normally from a 4-mm-thick plastic slab (density 1.14 g cm–3) with an energy...
A positron emerges normally from a 4-mm-thick plastic slab (density 1.14 g cm–3) with an energy of 1.62 MeV. What was its energy when it entered the slab?