Question

A person with lymphoma receives a dose of 45 Gy in the form of γ radiation during a course of radiotherapy. Most of this dose is absorbed in 18 g of cancerous lymphatic tissue. (a) How much energy is absorbed by the cancerous tissue? (b) If this treatment consists of five 20 minute sessions per week over the course of 5 weeks and just 1% of the γ photons in the γ ray beam are absorbed, what is the power of the γ ray beam? (c) If the γ ray beam consists of just 0.5% of the γ photons emitted by the γ source, each of which has energy of 0.03MeV, what is the activity (in Ci) of the γ ray source?

Answer 1

Answer:

Explanation:

a)  Energy absorbed by cancerous tissue E = R x M

R is the radiation dosage

M is the mass of the lymphatic

Energy absorbed by cancerous tissue

$=45\times18\times10^-^3J\\\\=810\times10^-^3J$

b) If only 1% of the total energy Et is absorbed over 20min session for 5 weeks = E

And the time of the period of the course is t = 20 x 5 = 100 min

E = 0.01Et

Total energy of gamma ray beam

Et = E x 100

= 810 x 10⁻³ x 100

= 8100J

Power of gamma ray beam is P

$P=\frac{Et}{t} \\\\=\frac{8100J}{100\times60} \\\\=0.0135W$

c) The total activity

$A=\frac{-dN}{dt} = dN decays/unit\ time$

The Number of decays dN = Es Total energy emitted at sourse per seconds / energy emitted per day

$dN=\frac{ Es}{ 0.03Mev}$

The Energy source Es per seconds = 0.05Ps

P= 0,0135W

The Power emitted at source

$Ps = \frac{0.0135}{0.05} \\\\=0.27W$

The Energy emitted per decay = 0.03MeV

$=0.03\times10^6\times1.602\times10^-^1^9J$

$A=\frac{-dN}{dt} \\\\=\frac{0.27}{0.03\times10^6\times1.602\times10^-^1^9} \\\\=\frac{0.27}{0.04806\times10^-^1^3} \\\\=5.618\times10^1^3/sec$