Question

the maximum background radiation that the average person receives in one year is about 3 msv. choose all the answers that are correct.
after 30 years, a person has received about the same amount of background radiation as they would after 30 mammograms (one mammogram a year from age 35 to 65 years of age).
if someone gets one mammogram a year, then when the radiation from this procedure is added to their maximum background radiation for the year, they have doubled the amount of radiation they received for that year.
a person would have to get 30 chest x - rays to get the same amount of radiation that they would get in one mammogram.
a 60 year old (who has only received background radiation) will have a lifetime radiation exposure that is equal to a 50 year old (who has had one mammogram a year for the past 10 years).
a person would have to be scanned at an airport security about 12,000 times to get the same amount of radiation that they receive naturally in one year

Explanation:

Step1: Determine radiation values

Assume mammogram is \(0.1\) mSv, chest - x ray is \(0.1\) mSv, airport scan is \(0.00025\) mSv.

Step2: Analyze each option

• Option 1:

Background in 30 years: \(3\times30 = 90\) mSv.
Mammograms (30): \(0.1\times30=3\) mSv. Not equal.

• Option 2:

If background is \(3\) mSv and mammogram is \(0.1\) mSv. \(3 + 0.1=3.1
eq6\) (doubled \(3\times2\)).

• Option 3:

One mammogram (\(0.1\) mSv). Chest - x ray (\(0.1\) mSv). \(0.1\div0.1 = 1
eq30\).

• Option 4:

60 - year - old background: \(3\times60=180\) mSv.
50 - year - old: \(3\times50+0.1\times10=150 + 1=151\) mSv. Not equal.

• Option 5:

Natural in one year (\(3\) mSv).
Number of scans: \(3\div0.00025=\frac{3}{\frac{25}{100000}}=\frac{3\times100000}{25}=12000\).

Answer:

Only the last option (A person would have to be scanned at an airport security about 12,000 times to get the same amount of radiation that they receive naturally in one year) is correct.