Question

At a distance of 41 ft , an ionizing radiation source delivers 5.0 rem of radiation. How close could you get to the source and still have no biological effects? Express your answer to two significant figures and include the appropriate units.

Answer 1

The student can get as close as infinity ft to the ionizing radiation source and still have no biological effects.

Step-by-step explanation:

To determine how close you can get to the ionizing radiation source without experiencing biological effects, we need to calculate the radiation dose at different distances using the inverse square law. The inverse square law states that the intensity of radiation decreases as the square of the distance increases.

Let's use the formula:

I1/I2 = (d2^2)/(d1^2)

Where I1 is the initial intensity, I2 is the final intensity, d1 is the initial distance, and d2 is the final distance.

Given that the initial intensity is 5.0 rem and the initial distance is 41 ft, we can rearrange the formula to solve for d2:

(I1/I2) = (d2^2)/(d1^2)

d2^2 = (d1^2) x (I1/I2)

d2 = sqrt((d1^2) x (I1/I2))

Substituting the values:

d2 = sqrt((41 ft)^2 x (5.0 rem/0 rem))

d2 = sqrt(1681 ft^2 x infinity)

d2 = infinity ft

According to the calculation, you could get as close as infinity ft to the source and still have no biological effects.

Answer 2

at distance 18.33 ft no biological effects

Step-by-step explanation:

given data

distance = 41 ft

radiation = 5 rem

to find out

how close get so no biological effects

solution

we know here that intensity of radiation R is inversely proportional to (radiation)²

so here equation will be

`(I1)/(I2)  = ((R2)/(R1))^(2)` .............1

here I is intensity

so here I1 = 25

because for 0 to 25 there is no detectable effects and for 25 to 100 it will . temporary decrease so

we take 25 because dose is less than 25 so we take that highest value

so

`(I1)/(I2)  = ((R2)/(R1))^(2)`

`(25)/(5)  = ((41)/(R1))^(2)`

R1 = 18.33 ft

so at distance 18.33 ft no biological effects