# Answer on Atomic and Nuclear Physics Question for James

Question #25858

1. A decommissioned nuclear power station still produces 0.5 MW of power five years after being turned off. Assuming that each decay releases 0.16 MeV of energy, calculate activity of the reactor in Kilo becquerels (KBq).

2. The ratio of carbon-14 to carbon-12 in a prehistoric wooden artifact is measured to be one-sixteenth of the ratio measured in a fresh sample of wood from the same region. The half life of carbon-14 is approximately 5,730 years. Determine its age.

3. What was the effective dose received by a cancer patient who received 2.8nSv to his lung and 350(mu)Sv to his colon. Compare your result with a normal background radiation and make and informed guess as to how likely the patient is to develop somatic effects after treatment

2. The ratio of carbon-14 to carbon-12 in a prehistoric wooden artifact is measured to be one-sixteenth of the ratio measured in a fresh sample of wood from the same region. The half life of carbon-14 is approximately 5,730 years. Determine its age.

3. What was the effective dose received by a cancer patient who received 2.8nSv to his lung and 350(mu)Sv to his colon. Compare your result with a normal background radiation and make and informed guess as to how likely the patient is to develop somatic effects after treatment

Expert's answer

1. A decommissioned nuclear power station still produces 0.5 MW of power five years after being turned off. Assuming that each decay releases 0.16 MeV of energy, calculate activity of the reactor in Kilo becquerels (KBq).

P = ΔE/Δt

P - power,

ΔE - energy, which station produces during time Δt

In another hand:

ΔE = e*A*Δt

e - energy of 1 decay

A - activity of the reactor

Therefore:

P = A*e*Δt/Δt = A*e

A = P/e = 0.5 MW / 0.16 MeV = 5*10^5 W / [1.6*10^5/(1.6*10^19) J ] = 5*10^19 (W/J)=5*10^19(W/(W*c))= 5*10^19 (1/c)

2. The ratio of carbon-14 to carbon-12 in a prehistoric wooden artifact is measured to be one-sixteenth of the ratio measured in a fresh sample of wood from the same region. The half life of carbon-14 is approximately 5,730 years. Determine its age.

n = n0*2^(-t/T)

n - concentration carbon-14 in artifact

n0 - concentration carbon-14 in a fresh sample of wood

T - the half life of carbon-14

t - age of wooden artifact

n/n0 = 1/16 = 2^(-t/T) = 2^(-4)

t/T = 4

t = 4*T = 4*5,730 years = 22920 years

3. What was the effective dose received by a cancer patient who received 2.8nSv to his lung and 350(mu)Sv to his colon. Compare your result with a normal background radiation and make and informed guess as to how likely the patient is to develop somatic effects after treatment

The effective dose in radiation protection and radiology is a measure of the cancer risk to a whole organism due to ionizing radiation delivered non-uniformly to part(s) of its body. It takes into account both the type of radiation and the nature of each organ being irradiated. More precisely, the effective dose of radiation (E) is found by calculating a weighted average of the eqiuvalent dose (HT) in different body tissues, with the weighting factors (WT) designed to reflect the different importance of tissue types to the danger to the whole organism.

E = W1*H1+W2*H2 = W1*2.8*10^(-9)Sv + W2*350*10^(-6)Sv

W1 = 0.12 - tissue weighting factor for lung;

W2 = 0.12 - tissue weighting factor for colon;

E = 0.12*2800*10^(-6) + 0.12*350*10^(-6) = 336*10^(-6) + 42*10^(-6) = 378 (mu)Sv

The worldwide average natural dose to humans is about 2.4 mSv per year = 2400 (mu)Sv per year

Somatic effects appear in the exposed person. Radiation effects are divided into somatic effects and genetic effects. The former affects the function of cells and organs, whereas the latter affects the future generations.

P = ΔE/Δt

P - power,

ΔE - energy, which station produces during time Δt

In another hand:

ΔE = e*A*Δt

e - energy of 1 decay

A - activity of the reactor

Therefore:

P = A*e*Δt/Δt = A*e

A = P/e = 0.5 MW / 0.16 MeV = 5*10^5 W / [1.6*10^5/(1.6*10^19) J ] = 5*10^19 (W/J)=5*10^19(W/(W*c))= 5*10^19 (1/c)

2. The ratio of carbon-14 to carbon-12 in a prehistoric wooden artifact is measured to be one-sixteenth of the ratio measured in a fresh sample of wood from the same region. The half life of carbon-14 is approximately 5,730 years. Determine its age.

n = n0*2^(-t/T)

n - concentration carbon-14 in artifact

n0 - concentration carbon-14 in a fresh sample of wood

T - the half life of carbon-14

t - age of wooden artifact

n/n0 = 1/16 = 2^(-t/T) = 2^(-4)

t/T = 4

t = 4*T = 4*5,730 years = 22920 years

3. What was the effective dose received by a cancer patient who received 2.8nSv to his lung and 350(mu)Sv to his colon. Compare your result with a normal background radiation and make and informed guess as to how likely the patient is to develop somatic effects after treatment

The effective dose in radiation protection and radiology is a measure of the cancer risk to a whole organism due to ionizing radiation delivered non-uniformly to part(s) of its body. It takes into account both the type of radiation and the nature of each organ being irradiated. More precisely, the effective dose of radiation (E) is found by calculating a weighted average of the eqiuvalent dose (HT) in different body tissues, with the weighting factors (WT) designed to reflect the different importance of tissue types to the danger to the whole organism.

E = W1*H1+W2*H2 = W1*2.8*10^(-9)Sv + W2*350*10^(-6)Sv

W1 = 0.12 - tissue weighting factor for lung;

W2 = 0.12 - tissue weighting factor for colon;

E = 0.12*2800*10^(-6) + 0.12*350*10^(-6) = 336*10^(-6) + 42*10^(-6) = 378 (mu)Sv

The worldwide average natural dose to humans is about 2.4 mSv per year = 2400 (mu)Sv per year

Somatic effects appear in the exposed person. Radiation effects are divided into somatic effects and genetic effects. The former affects the function of cells and organs, whereas the latter affects the future generations.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment