91Ó°ÊÓ

\([\mathrm{CP}\) An \(\mathrm{x}\)-ray tube is operating at voltage \(V\) and current \(I .\) (a) If only a fraction \(p\) of the electric power supplied is converted into \(x\) rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ), at what average rate would its temperature rise if there were no thermal losses? (c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at \(26 \mathrm{kV}\) and \(60.0 \mathrm{~mA}\) that converts \(1.0 \%\) of the electric power into \(x\) rays. Assume that the \(0.250-\mathrm{kg}\) target is made of lead \((c=130 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}) .\) (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

Step by step solution

01

Determine Power Supplied

The total power supplied to the x-ray tube can be calculated using the formula for electric power: \( P = V \times I \). For part (c), we use the given voltage \( V = 26 \text{kV} = 26,000 \text{V} \) and current \( I = 60.0 \text{mA} = 0.060 \text{A} \). Therefore, the power \( P \) is: \[ P = 26,000 \times 0.060 = 1560 \text{W} \].

02

Calculate Power Converted to X-rays

Since only a fraction \( p \) of the total power is converted into x-rays, the power converted to x-rays is \( P_x = p \times P \). For part (c), \( p = 0.01 \), so \( P_x = 0.01 \times 1560 = 15.6 \text{W} \).

03

Calculate Power Delivered to Target

The rate of energy delivery to the target (not converted to x-rays) is the remaining part of the power. It is given by \( P_{target} = P - P_x \). \[ P_{target} = 1560 - 15.6 = 1544.4 \text{W} \].

04

Calculate Temperature Rise Rate of Target

Using the formula for energy and specific heat, the rate of temperature rise can be calculated as \( \frac{dT}{dt} = \frac{P_{target}}{m \times c} \). With \( m = 0.250 \text{ kg} \) and \( c = 130 \text{ J/kg} \cdot \text{K} \), we get: \[\frac{dT}{dt} = \frac{1544.4}{0.250 \times 130} = \frac{1544.4}{32.5} \approx 47.5 \text{ K/s} \].

05

Consider Suitable Target Materials

A practical target material should have a high melting point, good thermal conductivity, and sufficient atomic number to efficiently generate x-rays. Suitable elements include tungsten, molybdenum, and platinum, which have high melting points and desirable atomic structures for x-ray production.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Power Conversion

In an X-ray tube, electric power is converted into X-ray radiation. The process begins with the power supplied to the tube, defined by the formula, \( P = V \times I \), where \( V \) is the voltage and \( I \) is the current. For instance, if an X-ray tube operates at 26 kV and 60 mA, the power supplied is calculated as \( P = 26,000 \times 0.060 = 1560 \text{ W} \).

Notably, only a small fraction of this electric power, denoted by \( p \), is converted into X-ray radiation. If \( p \) is 1%, the power converted to X-rays is \( 15.6 \text{ W} \). This signifies that the efficiency of power conversion to X-rays is relatively low. Most of the power is dissipated as heat.

Specific Heat

The specific heat of a material is a measure of the amount of heat required to raise the temperature of a unit mass by one degree Celsius (or one Kelvin). It is an inherent property of materials, denoted by \( c \) and measured in \( \text{J/kg} \cdot \text{K} \).

In the context of an X-ray tube, the target material's specific heat is crucial, as it affects how the material handles excess heat generated during operation. For example, lead, a common target material, has a specific heat of \( 130 \text{ J/kg} \cdot \text{K} \). Low specific heat means the material heats up quickly, which can be challenging for managing heat in X-ray generation.

Temperature Rise Rate

The temperature rise rate of a material is indicative of how quickly it heats up when energy is applied. It can be determined by the formula \( \frac{dT}{dt} = \frac{P_{target}}{m \times c} \). Here, \( P_{target} \) represents the power delivered to the target, \( m \) is the mass, and \( c \) is the specific heat.

Taking our previous example, with a power of \( 1544.4 \text{ W} \) delivered to a 0.250-kg lead target, the temperature rise rate is calculated as \( \frac{1544.4}{0.250 \times 130} = 47.5 \text{ K/s} \). This high rate indicates how important it is to manage the heat effectively during the operation of an X-ray tube.

Suitable Target Materials

Choosing suitable materials for the target in an X-ray tube is crucial for efficient and safe operation. Practical target materials should possess several properties:

• High melting point: To withstand the intense heat generated during operation.
• Good thermal conductivity: To quickly dissipate heat and prevent overheating.
• High atomic number: To improve the efficiency of X-ray production.

Commonly used materials include tungsten, due to its high melting point and density, molybdenum, known for its thermal management capabilities, and platinum, which also possesses good mechanical properties and high atomic number.

X-ray Production Efficiency

X-ray production efficiency in tubes is a measure of how well the supplied electric power is transformed into X-ray radiation. Typically, only a small percentage, often around 1% or less, of the electric power is actually converted into useful X-rays, as seen in the example calculation where only 15.6 W out of 1560 W contributes to X-rays.

This low efficiency is usually due to the significant energy loss in the form of heat. Enhanced design and material selection, such as using targets with high atomic numbers, can somewhat improve efficiency by increasing the likelihood of X-ray production during each interaction with electrons.