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An 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 \(\mathbf{x}\) . rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat capacity \(c(\text { in } \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}),\) 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 18.0 \(\mathrm{kV}\) and 60.0 \(\mathrm{mA}\) that converts 1.0\(\%\) of the electric power into \(\mathrm{x}\) 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

Calculate the Electric Power Supplied

The electric power supplied to the x-ray tube can be calculated using the formula for electric power, \( P = V \times I \), where \( V \) is the voltage and \( I \) is the current. Plug in the given values of voltage and current to calculate \( P \).For \( V = 18.0 \text{ kV} = 18000 \text{ V} \) and \( I = 60.0 \text{ mA} = 0.060 \text{ A} \):\[ P = 18000 \, \text{V} \times 0.060 \, \text{A} = 1080 \, \text{W} \]

02

Calculate Energy Delivered to the Target

Given that only a fraction \( p \) of the electric power is converted into x-rays, the rate at which energy is delivered to the target is given by \( P_t = (1 - p) \times P \). Here, \( p = 0.01 \) or 1%.\[ P_t = (1 - 0.01) \times 1080 \, \text{W} = 0.99 \times 1080 \, \text{W} = 1069.2 \, \text{W} \]

03

Calculate Temperature Rise Rate

The rate of temperature rise \( \frac{\Delta T}{\Delta t} \) is determined using the formula \( P_t = m \times c \times \frac{\Delta T}{\Delta t} \), where \( m = 0.250 \, \text{kg} \) and \( c = 130 \, \text{J/kg} \cdot \text{K} \).Rearranging to find \( \frac{\Delta T}{\Delta t} \):\[ \frac{\Delta T}{\Delta t} = \frac{P_t}{m \times c} = \frac{1069.2 \, \text{W}}{0.250 \, \text{kg} \times 130 \, \text{J/kg} \cdot \text{K}} = \frac{1069.2}{32.5} \approx 32.89 \, \text{K/s} \]

04

Discuss Suitable Target Materials

A practical target material must have a high melting point to withstand the heat, a high atomic number to efficiently produce x-rays, and good thermal conductivity to dissipate heat. Suitable target elements include tungsten and molybdenum due to their high melting points and efficacy in 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

Electric power is the rate at which electrical energy is transferred by an electric circuit. In the case of an x-ray tube, electric power determines how much energy is available to generate x-rays. It is calculated using the formula:

• \( P = V \times I \)

where \( V \) is the voltage and \( I \) is the current. For example, when the voltage is 18,000 V (or 18.0 kV) and the current is 0.060 A (or 60.0 mA), the electric power supplied to the x-ray tube is:

• \( P = 18,000 \, \text{V} \times 0.060 \, \text{A} = 1,080 \, \text{W} \)

This power is not entirely converted into x-rays, as typically only a small fraction \( p \) is used, and the rest contributes to heating the target.

Specific Heat Capacity

Specific heat capacity is a property of a material that describes how much energy is needed to raise the temperature of a given mass by one degree Celsius (or one Kelvin). It is denoted by \( c \) and has units of \( \text{J/kg} \cdot \text{K} \). This concept is crucial when examining how the target in an x-ray tube heats up.

• Materials with high specific heat capacities absorb more energy without a significant change in temperature.
• Conversely, materials with low specific heat capacities will heat up quickly.

For instance, in the given exercise, the lead target has a specific heat capacity of 130 \( \text{J/kg} \cdot \text{K} \). This means that for every 1 kg of lead, 130 Joules are needed to increase the temperature by 1 Kelvin. When the electric power contributes to heating the target, its temperature increase rate can be calculated as:

• \[ \frac{\Delta T}{\Delta t} = \frac{P_t}{m \times c} \]

where \( P_t \) is the power causing the temperature rise, \( m \) is the mass, and \( c \) is the specific heat capacity.

Thermal Conductivity

Thermal conductivity is a measure of how well a material can conduct heat. In the context of an x-ray tube, the target material must efficiently dissipate heat to prevent overheating. This property directly affects the operational efficiency and durability of the x-ray tube.

• Materials with high thermal conductivity can rapidly transfer heat away from the point of origin, thus preventing damage.
• Conversely, materials with low thermal conductivity retain heat, which could result in damage over time due to excessive temperatures.

For efficient cooling in x-ray tubes, thermal conductivity plays a complementary role alongside other properties such as specific heat capacity and melting point. It is essential that these properties are balanced in the choice of target material to ensure effective operation and longevity.

Target Material Properties

The target material in an x-ray tube must possess several key properties to function effectively. These properties ensure that the tube operates efficiently and safely:

• **High Melting Point:** The target needs to withstand high temperatures without melting.
• **High Atomic Number:** Helps in efficiently producing x-rays by maximizing the likelihood of interactions that generate x-rays.
• **Good Thermal Conductivity:** Ensures efficient heat dissipation to prevent overheating and preserve the target's integrity.

Some common materials used as targets include tungsten and molybdenum, known for their high melting points and ability to efficiently create x-rays. These metals balance the need for durability, efficient heat transfer, and effective x-ray production.