Assume that the risk of ionizing radiation causing a cancer is 8 x 10^-2/person

Question

Assume that the risk of ionizing radiation causing a cancer is 8 x 10^-2/person Gy and that this risk applies to low doses and low dose rate of ionising radiation. Assume also that the average amount of background radiation each person receives in Canada s 3.2 mGy/year and that the population of Canada is 31.5 x 10^6 persons. How many cancers per year are coursed by background radiation? If a person lives in a region where the background radiation is 7.5 mGy/year, what is the risk she will have a cancer caused by radiation by time she is 84 years old?

Explanation / Answer

We calculate the fraction of cancers due to background radiation

remenber Gy= 103 mGy

8 10-2 Gy------   100 probability

3.2 10-3 Gy-----     X

X= (3.2 10-3 /8 10-2 ) 100

X=4%

People with cancer will be the population by the probability

Population* probability/100

31.5 106 0.04=   1.26 106 personas with cancer

Second part

Rad = 7.5 10-3 Gy/ year

T = 84 years

Probability calculus

  8 10-2   ------   100

7.5 10-3   -----     X

X= (7.5 10-3 /8 10-2 )100 = 9.375%

If the person lived life in this place all time the probability increases linearly with age since the dose of radiation is cumulative, assuming that elimination mechanisms there are no by the human body.

9.375% * 84 = 787%

This result has no true, since the mechanisms of elimination of radiation by the body are varied, for example: cell renewal, etc. As a result the probability remains stable over time 9.375%

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