Problem 78
Question
The intensity of \(\mathrm{X}\)-ray radiation depends on the distance from the source of radiation. If the distance from the source of radiation changes from \(D_{1}\) to \(D_{2}\), the intensity changes from intensity \(I_{1}\) to \(I_{2}\). The formula for finding the new intensity is \(I_{2}=\frac{I_{1} \cdot\left(D_{1}\right)^{2}}{\left(D_{2}\right)^{2}}\). When \(D_{1}\) is \(25 \mathrm{~m}\), the intensity \(I_{1}\) is 620 roentgen per hour. Find the intensity of radiation \(I_{2}\) if \(D_{2}\) is \(5 \mathrm{~m}\).
Step-by-Step Solution
Verified Answer
The final intensity \(I_2\) is 15500 roentgen per hour.
1Step 1 - Write Down Given Values
Identify the given values from the problem. Here, they are: Distance from the source at initial point: \[ D_1 = 25 \text{ m} \] Initial intensity: \[I_1 = 620 \text{ roentgen per hour}\] Distance from the source at final point: \[ D_2 = 5 \text{ m} \]
2Step 2 - Write Down the Formula
The next step is to write down the formula provided in the problem for finding the new intensity. It is:\[ I_2 = \frac{I_1 \cdot (D_1)^2}{(D_2)^2} \]
3Step 3 - Substitute the Values
Substitute the given values into the formula. It becomes: \[ I_2 = \frac{620 \cdot (25)^2}{(5)^2} \]
4Step 4 - Simplify the Equation
Next, simplify the right side of the equation step-by-step:First, calculate \[ (25)^2 = 625 \]Then, calculate \[ (5)^2 = 25 \]Update the equation with these values: \[ I_2 = \frac{620 \cdot 625}{25} \]
5Step 5 - Perform the Multiplication and Division
Perform the multiplication and division to find the final value:\[ I_2 = \frac{620 \cdot 625}{25} = 620 \cdot 25 \]Calculate \[ 620 \cdot 25 = 15500 \]
6Step 6: Final Step - State the Solution
The final intensity of radiation is:\[ I_2 = 15500 \text{ roentgen per hour} \]
Key Concepts
intensity of radiationdistance and intensity relationshipalgebraic manipulation
intensity of radiation
The intensity of radiation measures how much energy passes through a certain area over a specified time period. For example, in the given problem, the intensity is measured in roentgen per hour (R/hr), which tells us how much X-ray radiation energy is received every hour. Understanding radiation intensity is crucial in fields like radiology and nuclear physics, as it helps determine safe exposure levels and the effectiveness of radiation-based treatments. Remember, greater intensity means more energy and potentially higher impact on materials and living tissues.
distance and intensity relationship
The relationship between distance from a radiation source and its intensity is described by the inverse square law. This law states that the intensity of radiation is inversely proportional to the square of the distance from the source. In simple terms, when you double the distance from the source, the radiation intensity drops to a quarter of its initial value. This principle is key in the exercise you are working on. Given the formula: \[ I_2 = \frac{I_1 \cdot (D_1)^2}{(D_2)^2} \] You can see how the intensity (\(I_2\)) changes based on distances \(D_1\) and \(D_2\). For example, if \(D_1\) increases and \(D_2\) remains constant, the intensity \(I_2\) will decrease. This formula is a practical application of the inverse square law, helping us to predict changes in intensity with varying distances.
algebraic manipulation
Algebraic manipulation involves rearranging and simplifying equations to solve for unknown variables. In our exercise, we start with the formula: \[ I_2 = \frac{I_1 \cdot (D_1)^2}{(D_2)^2} \] By substituting the known values (\(D_1 = 25 \text{ m}\), \(I_1 = 620 \text{ roentgen per hour}\), and \(D_2 = 5 \text{ m}\)) into the formula, we get: \[ I_2 = \frac{620 \times (25)^2}{(5)^2} \] Next, we simplify the right side of the equation step-by-step:
- First, calculate \( (25)^2 = 625 \).
- Then, calculate \( (5)^2 = 25 \).
- Finally, update the formula to \( I_2 = \frac{620 \times 625}{25} \).
Other exercises in this chapter
Problem 77
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\(\left(\frac{2}{3}+\frac{1}{2}\right)^{2}\)
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\(18\left(\frac{5}{6} h-\frac{1}{3}\right)-7\)
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