A vial containing 32 millicuries is emitting radiation at a rate of 1 roentgen per hour at a distance of 1 meter. If a technologist stands at a distance of 2 meters from the vial, how much radiation (in roentgens) does the technologist receive during a 15-minute period?

To calculate the dose received by a technologist standing at a certain distance from a radioactive source, you can use the inverse square law and the exposure rate provided.

The inverse square law states that the intensity of radiation is inversely proportional to the square of the distance from the source. The formula for the inverse square law is:

${\ufffd}_{2}=\frac{{\ufffd}_{1}\cdot ({\ufffd}_{1}{)}^{2}}{({\ufffd}_{2}{)}^{2}}$

Where:

- ${\ufffd}_{2}$ is the new intensity at the second distance,
- ${\ufffd}_{1}$ is the initial intensity at the first distance,
- ${\ufffd}_{1}$ is the initial distance,
- ${\ufffd}_{2}$ is the second distance.

Given that the vial emits 1 roentgen per hour at 1 meter (${\ufffd}_{1}=1\text{\hspace{0.17em}}\text{R/hr}$ at ${\ufffd}_{1}=1\text{\hspace{0.17em}}\text{m}$), we want to find the intensity at 2 meters (${\ufffd}_{2}=2\text{\hspace{0.17em}}\text{m}$).

${\ufffd}_{2}=\frac{1\cdot (1{)}^{2}}{(2{)}^{2}}=\frac{1}{4}\text{\hspace{0.17em}}\text{R/hr}$

Now, we know the new intensity at 2 meters is $\frac{1}{4}\text{\hspace{0.17em}}\text{R/hr}$. To find the dose received by the technologist in 15 minutes ($\frac{1}{4}$ of an hour), you can simply calculate:

$\text{Dose}=\text{Intensity}\times \text{Time}$

$\text{Dose}=\frac{1}{4}\text{\hspace{0.17em}}\text{R/hr}\times \frac{1}{4}\text{\hspace{0.17em}}\text{hr}=\frac{1}{16}\text{\hspace{0.17em}}\text{roentgen}$

So, the technologist would receive $\frac{1}{16}$ roentgen during the 15 minutes standing at a distance of 2 meters from the vial.

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