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?
Can you answer this please !
Doubling the distance will result in a count of (1/2)^2= 1/4 the original
count. 1/4 of 1000 = 250 counts per minute.
To determine the number of counts per second when the detector is 2 meters from the sample, we can use the inverse square law of physics, which states that the intensity of radiation is inversely proportional to the square of the distance.
Let's denote the initial counts per second as C1 and the initial distance as D1. We are given that C1 = 1000 counts per second and D1 = 1 meter.
Let's also denote the new counts per second as C2 and the new distance as D2. We need to find C2 when D2 = 2 meters.
According to the inverse square law, the intensity of radiation is given by:
I1 / I2 = (D2 / D1)^2
Substituting the given values:
1000 / C2 = (2 / 1)^2
Simplifying the equation:
1000 / C2 = 4
To isolate C2, we can rewrite the equation as:
C2 = 1000 / 4
C2 = 250 counts per second
Therefore, when the detector is 2 meters from the sample, the number of counts per second observed would be 250.
To find out how many counts per second would be observed when the detector is 2 meters from the sample, we can use the inverse square law for radiation. This law states that the intensity of radiation decreases with the square of the distance.
First, let's set up the equation:
I1 / I2 = (D2 / D1)^2
Where:
I1 is the intensity of radiation at the initial distance (1 meter)
I2 is the intensity of radiation at the new distance (2 meters)
D1 is the initial distance (1 meter)
D2 is the new distance (2 meters)
We are given that I1 = 1000 counts per second and D1 = 1 meter. We want to find I2.
Rearranging the equation to solve for I2, we have:
I2 = I1 * (D1 / D2)^2
Substituting the given values, we get:
I2 = 1000 * (1 / 2)^2 = 1000 * (1/4) = 250 counts per second
Therefore, when the detector is 2 meters from the sample, the observer would observe 250 counts per second.