A60co source with activity 2.4×10−4 ci is imbedded in a tumor that has mass 0.550 kg . the source emits γ photons with average energy 1.25 mev. half the photons are absorbed in the tumor, and half escape.

a. what energy is delivered to the tumor per second?

b. what absorbed dose (in rad) is delivered per second?

c. what equivalent dose (in rem) is delivered per second if the rbe for these γ rays is 0.70?

d. what exposure time is required for an equivalent dose of 180 rem ?

(A). The energy delivered to the tumor per second is .

(B). The absorbed dose delivered per second is

(C). The equivalent dose delivered per second is

(D). The exposure time is 18 days.

Explanation:

Given that,

Activity of the source

Mass of tumor = 0.550 kg

Average energy = 1.25 MeV

Half the photons are absorbed in the tumor, and half escape

1 Gy =100 rad

(A). We need to calculate the energy delivered to the tumor per second

Using formula of energy delivered

Put the value into the formula

(B). We need to calculate the absorbed dose delivered per second

Using formula of absorbed dose

Put the value into the formula

(C). If the RBE for these γ rays is 0.70

We need to calculate the equivalent dose delivered per second

Using formula of  the equivalent dose

Put the value into the formula

(D). We need to calculate the exposure time for required for an equivalent dose of 180 rem

Using formula of exposure time

Hence, (A). The energy delivered to the tumor per second is .

(B). The absorbed dose delivered per second is

(C). The equivalent dose delivered per second is

(D). The exposure time is 18 days.

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explanation:

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).