Problem 38 - University Physics with Modern Physics

Question

CP 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 \(x\) rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in \(J / k g \cdot K ),\) 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

Answer

(a) 1069.2 W; (b) 32.9 K/s; (c) Calculate using given values; (d) High melting point, thermal conductivity; tungsten, molybdenum.

Step 1: Determine the Total Electric Power

The power supplied to the x-ray tube is the product of voltage and current. Thus, the electric power \( P \) is given by:\[P = V \cdot I\]Here, both \( V \) and \( I \) are given, so substitute their values for calculation.

Step 2: Calculate the Rate of Energy Delivered to the Target

From the given fraction \( p \) that converts into x-rays, determine the power actually delivered to the target:\[P_{target} = (1 - p) \cdot P\]Using the values of \( P \) from Step 1, calculate \( P_{target} \).

Step 3: Calculate the Average Rate of Temperature Increase

The rate at which the temperature of the target increases is found using the specific heat formula:\[\frac{dT}{dt} = \frac{P_{target}}{m \cdot c}\]where:- \( P_{target} \) is from Step 2,- \( m \) is the mass of the target,- \( c \) is the specific heat capacity of the target material.Substitute to find \( \frac{dT}{dt} \).

Step 4: Numerical Evaluation

Substitute the given values to evaluate expressions found in Steps 1-3:- \( V = 18.0 \: \mathrm{kV} = 18000 \: \mathrm{V} \)- \( I = 60.0 \: \mathrm{mA} = 0.060 \: \mathrm{A} \)- \( p = 0.01 \)- \( m = 0.250 \: \mathrm{kg} \)- \( c = 130 \: \mathrm{J/kg \cdot K} \)Using these values, calculate:\( P = 18000 \times 0.060 = 1080 \: \mathrm{W} \)\( P_{target} = (1 - 0.01) \times 1080 = 1069.2 \: \mathrm{W} \)\( \frac{dT}{dt} = \frac{1069.2}{0.250 \times 130} \approx 32.9 \: \mathrm{K/s} \)

Step 5: Suitable Characteristics for Target Material

For a practical x-ray tube target, the material should have high melting point to withstand heating, high atomic number for efficient x-ray production, and good thermal conductivity to manage heat. Suitable target elements could include tungsten (often used for its high melting point and atomic number) or molybdenum.

Key Concepts

Electric Power Calculation

In an x-ray tube, it is important to understand how electric power plays a role in generating x-rays. The electric power (\(P\)) supplied to the tube can be calculated using the formula: \[P = V \cdot I\] where \(V\) is the voltage and \(I\) is the current flowing through the tube.
This calculation helps determine how much energy is being used by the x-ray tube. For example, if the voltage is 18,000 volts and the current is 0.060 amperes, the electric power would be \(1080 \, ext{W}\).
However, not all this power is used for x-ray production. Only a fraction \(p\) is converted into x-rays, while the rest heats the target material.

Specific Heat Capacity

Specific heat capacity is a property that helps us understand how a material responds to added heat. It is defined as the amount of heat required to raise the temperature of one kilogram of a substance by one degree Kelvin (or Celsius).
In the context of an x-ray tube, we use this concept to calculate the rate at which the target material's temperature increases. The formula to find the rate of temperature increase (\(\frac{dT}{dt}\)) is: \[\frac{dT}{dt} = \frac{P_{target}}{m \cdot c}\]where:

• \(P_{target}\) represents the power delivered to the target,
• \(m\) is the mass of the target, and
• \(c\) is the specific heat capacity of the material.

For instance, if the target mass is 0.250 kg and the specific heat capacity is 130 J/kg·K, you can calculate how fast the temperature of the lead target increases when exposed to unsuppressed energy.

Thermal Properties of Materials

The thermal properties of materials are essential when considering the design and efficiency of x-ray tubes. These properties tell us how materials handle heat, and they include thermal conductivity, heat capacity, and melting point.
A good x-ray tube target must efficiently conduct heat away to prevent damage from high temperatures induced during x-ray production. Materials with high thermal conductivity are favorable because they distribute heat quickly and minimize temperature ripples.
Furthermore, a high heat capacity and melting point are important, so the target doesn’t melt or deteriorate under prolonged exposure. These properties help extend the lifespan of the x-ray tube and maintain stable operating conditions.

Target Material Characteristics

When it comes to selecting materials for the x-ray tube target, certain characteristics are crucial for effective x-ray production. A suitable target material generally has:

• A high atomic number, which enhances x-ray production efficiency.
• A high melting point, allowing it to withstand the considerable heat developed during x-ray emission.
• Good thermal conductivity to dissipate heat and prevent damage.

Tungsten is a common choice due to its remarkable atomic number and melting point. Molybdenum is another option, especially used in mammography, benefiting from its suitable x-ray production qualities at lower energy levels. These materials ensure that the x-ray tube functions efficiently, safely, and for a longer period.