If the gamma-ray dose rate is 50 mSv hr-1 at a distance of 1.0 m from a source, what is the dose rate at a distance of 10 m from the source?
If the gamma-ray dose rate is 50 mSv hr-1 at a distance of 1.0 m from a source, what is the dose rate at a distance of 10 m from the source?
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![The text presents a question regarding the calculation of gamma-ray dose rates at varying distances:
"If the gamma-ray dose rate is 50 mSv/hr at a distance of 1.0 m from a source, what is the dose rate at a distance of 10 m from the source?"
### Explanation:
To solve this, you can use the inverse square law, which states that the intensity of radiation is inversely proportional to the square of the distance from the source. This means:
\[ I_1 \times d_1^2 = I_2 \times d_2^2 \]
Where:
- \( I_1 \) is the initial dose rate (50 mSv/hr at 1.0 m)
- \( d_1 \) is the initial distance (1.0 m)
- \( I_2 \) is the dose rate at the new distance
- \( d_2 \) is the new distance (10 m)
Given that:
\[ 50 \, \text{mSv/hr} \times (1.0 \, \text{m})^2 = I_2 \times (10 \, \text{m})^2 \]
\[ 50 = I_2 \times 100 \]
\[ I_2 = \frac{50}{100} \]
\[ I_2 = 0.5 \, \text{mSv/hr} \]
Thus, the dose rate at a distance of 10 m from the source is 0.5 mSv/hr.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03c32ca7-a243-40d2-963f-bc8ec9cda4bb%2F998010b2-a600-4565-b410-6b376ad55a10%2F7bcy454_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The text presents a question regarding the calculation of gamma-ray dose rates at varying distances:
"If the gamma-ray dose rate is 50 mSv/hr at a distance of 1.0 m from a source, what is the dose rate at a distance of 10 m from the source?"
### Explanation:
To solve this, you can use the inverse square law, which states that the intensity of radiation is inversely proportional to the square of the distance from the source. This means:
\[ I_1 \times d_1^2 = I_2 \times d_2^2 \]
Where:
- \( I_1 \) is the initial dose rate (50 mSv/hr at 1.0 m)
- \( d_1 \) is the initial distance (1.0 m)
- \( I_2 \) is the dose rate at the new distance
- \( d_2 \) is the new distance (10 m)
Given that:
\[ 50 \, \text{mSv/hr} \times (1.0 \, \text{m})^2 = I_2 \times (10 \, \text{m})^2 \]
\[ 50 = I_2 \times 100 \]
\[ I_2 = \frac{50}{100} \]
\[ I_2 = 0.5 \, \text{mSv/hr} \]
Thus, the dose rate at a distance of 10 m from the source is 0.5 mSv/hr.
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