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Ch.11 Nuclear Chemistry
McMurry - Fundamentals of General, Organic, and Biological Chemistry 8th Edition
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 12

A β-emitting radiation source gives 250 units of radiation at a distance of 4.0 m. At what distance does the radiation drop to one-tenth its original value?

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Understand the problem: The intensity of radiation follows the inverse square law, which states that the intensity of radiation is inversely proportional to the square of the distance from the source. Mathematically, this can be expressed as \( I \propto \frac{1}{d^2} \), where \( I \) is the intensity and \( d \) is the distance.
Set up the relationship: Let the initial intensity \( I_1 \) be 250 units at a distance \( d_1 = 4.0 \, \text{m} \). The final intensity \( I_2 \) is one-tenth of the original intensity, so \( I_2 = \(\frac{I_1}{10}\) = 25 \, \(\text{units}\).
Use the inverse square law formula: \( \frac{I_1}{I_2} = \left( \frac{d_2}{d_1} \right)^2 \). Substitute \( I_1 = 250 \), \( I_2 = 25 \), and \( d_1 = 4.0 \, \text{m} \) into the equation.
Solve for \( d_2 \): Rearrange the formula to isolate \( d_2 \): \( d_2 = d_1 \sqrt{\frac{I_1}{I_2}} \). Substitute the known values into the equation.
Perform the calculation: Calculate \( d_2 \) using the values substituted in the previous step to determine the distance at which the radiation drops to one-tenth its original value.

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Inverse Square Law

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