How much energy is imparted to one cell during one day's treatment? Assume that the specific gravity of the tumor is 1 and that 1 J = 6 $\times$ 10$^{18}$ eV. (a) 120 keV; (b) 12 MeV; (c) 120 MeV; (d) 120 $\times$ 10$^3$ MeV.
Added by Christopher G.
Step 1
Step 1: Calculate the energy imparted to one cell using the formula: energy on one cell = energy given per day * density of the tumor * number of cells per cubic meter. Show more…
Show all steps
Your feedback will help us improve your experience
Dominador Tan and 85 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
How much energy is imparted to one cell during one day's treatment? Assume that the specific gravity of the tumor is 1 and that $1 \mathrm{~J}=6 \times 10^{18} \mathrm{eV}$ (a) $120 \mathrm{keV}$; (b) $12 \mathrm{MeV}$ (c) $120 \mathrm{MeV}$ (d) $120 \times 10^{3} \mathrm{MeV}$
(II) A cancer patient is undergoing radiation therapy in which protons with an energy of 1.2 MeV are incident on a 0.25 -kg tumor. (a) If the patient receives an effective dose of 1.0 rem, what is the absorbed dose? (b) How many proton are absorbed by the tumor? Assume OF $\approx 1 .$
Suppose the answer to problem 71 is 12 MeV/cell. How many Compton interactions will occur per cell in a single day's treatment? A. $120 \times 10^{6}$ B. $120 \times 10^{4}$ C. $120 \times 10^{2}$ D. 12
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD