A 3.0-kg tumor is being irradiated by a radioactive source. The tumor receives a

Question

A 3.0-kg tumor is being irradiated by a radioactive source. The tumor receives an absorbed dose of 14 Gy in a time of 700 s. Each disintegration of the radioactive source produces a particle that enters the tumor and delivers an energy of0.605MeV. What is the activity of the radioactive source? 6) The thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. The wavelength of 575 nm is reflected the most strongly from the film. What is the minimum thickness of the film?

Explanation / Answer

Given :

A) mass of tumor = 3 kg

time t =700 s

14 Gy

Energy E = 0.605Mev

As we know from formula,

Grays = energy / mass -------------------------------------------------(i)

energy = decays * 0.605MeV --------------------------------------------------(ii)

Activity = decays / time

decays = activity * time    ---------------------------------------------------(iii)      

first will subtract equ (iii) from equ (ii) and then putting value of energy into equ (i)

Wil get , Grays = activity * time * 0.605MeV / mass

= activity = Grays * mass / (time * 0.605MeV)

  1MeV=1.6x10^-13J

activity = 14 * 3.0 / (700 * 0.605 * 1.6x10^-13) =

= 6.198 * 10^11 -----------------------------------------------------Answer

B)

Strong reflection is referring to constructive interference between rays reflected off the top

phase shift of ½ for rays reflected off the top of the film (reflection against a more dense medium)

he minimum thickness (t) for a path diff. of ½ wave is given by ..
2 t = ½ ' .. (' = wavelength in film = air / n)

t = ¼( / n)= 575 nm / ( 1.5 * 4) = 95.83 nm ----------------Answer

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