If a sample emits 1000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 2 meters from the sample?
To answer this question, we can use the inverse square law, which states that the intensity of radiation decreases in proportion to the square of the distance from the source.
To apply the inverse square law to this problem, we need to find the ratio of intensities at two different distances.
The intensity of radiation can be thought of as the number of counts per second. Let's call the initial intensity at 1 meter I1, and the intensity at 2 meters I2.
According to the inverse square law:
(I1 / I2) = (d2 / d1)^2
where d1 is the initial distance (1 meter) and d2 is the new distance (2 meters).
Substituting the given values into the equation:
(I1 / I2) = (2 / 1)^2
Simplifying:
(I1 / I2) = 4
To find the intensity at 2 meters (I2), we can rearrange the equation:
I2 = I1 / 4
Given that the initial intensity (I1) is 1000 counts per second:
I2 = 1000 / 4 = 250 counts per second
Therefore, if the detector is 2 meters from the sample, approximately 250 counts per second would be observed.