Problem 48 - Master Resource Book in JEE Main Physics

Question

An X-ray tube is operating at \(50 \mathrm{kV}\) and \(20 \mathrm{~mA}\). The target material of the tube has a mass of \(1 \mathrm{mg}\) and specific heat \(495 \mathrm{~J} \mathrm{~kg}^{-1}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}^{-1} .\) One per cent of the supplied electric power is converted into \(X\)-rays and the entire remaining energy goes into heating the target. Then (a) a suitable target material must have a high the melting temperature (b) a suitable target material must have low thermal conductivity (c) the average rate of use of temperature of target would be \(2^{\circ} \mathrm{C} / \mathrm{s}\) (d) the minimum wavelength of the \(X\)-rays emitted is about \(0.25 \times 10^{-10} \mathrm{~m}\)

Step-by-Step Solution

Answer

Correct options are (a) and (d).

Step 1: Calculating Electric Power Supplied

The electric power supplied to the X-ray tube is calculated using the formula \( P = V imes I \), where \( V = 50\, \text{kV} = 50,000\, \text{V} \) and \( I = 20\, \text{mA} = 0.02\, \text{A} \). Therefore, \( P = 50,000 \times 0.02 = 1000 \text{ W} \).

Step 2: Determining Power Used for Heating

Since only 1% of the power is converted into X-rays, 99% of the power is used for heating.\[ P_{heat} = 0.99 \times 1000 = 990 \text{ W} \]

Step 3: Calculating Rate of Temperature Rise

The rate of temperature rise is given by \( q = mc\Delta T \), where \( \Delta T/\Delta t = P_{heat} / (mc) \). \( m = 1 \text{ mg} = 1 \times 10^{-6} \text{ kg} \), and \( c = 495 \text{ J} \text{ kg}^{-1}{\underline{\phantom{xx}}}^{\circ} \text{C}^{-1} \). Thus, \[ \Delta T/\Delta t = \frac{990}{1 \times 10^{-6} \times 495} = 2 \times 10^{6} \text{ C/s} \] But given options (c) asserts \( \Delta T/\Delta t = 2^{\circ} \text{C/s},\) which is not correct according to this calculation.

Step 4: Computing Minimum X-ray Wavelength

The minimum wavelength \( \lambda_{min} \) of produced X-rays is calculated using the formula \( \lambda_{min} = \frac{hc}{eV} \) where \( h = 6.63 \times 10^{-34} \text{ J.s} \), \( e = 1.602 \times 10^{-19} \text{ C} \), \( c = 3 \times 10^{8} \text{ m/s} \), \( V = 50,000 \text{ V} \). This gives: \[ \lambda_{min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.602 \times 10^{-19} \times 50,000} \approx 0.248 \times 10^{-10} \text{ m} \] Hence option (d) is approximately correct.

Step 5: Evaluate Suitability of Material

Suitable materials for X-ray tubes must have a high melting temperature to withstand high energy without melting, thus option (a) is correct. Additionally, a low thermal conductivity is not ideal since it would not allow efficient heat dissipation, so option (b) is incorrect.

Key Concepts

Electric Power Calculation

In X-ray tube physics, calculating electric power is fundamental to understanding the tube's operation. The concept revolves around determining the power input required to generate X-rays. This is calculated using the basic formula for electric power: \( P = V \times I \), where \( V \) is the voltage and \( I \) is the current.

For our specific example, the X-ray tube is operating at \( 50 \text{kV} \) (or \( 50,000 \text{V} \)) and \( 20 \text{mA} \) (or \( 0.02 \text{A} \)). Plugging these numbers into the formula gives:

- \( P = 50,000 \times 0.02 = 1000 \text{ W} \)

This means the tube consumes 1000 Watts of electric power. Understanding this allows us to delve deeper into how that energy gets distributed, particularly how much is used for producing X-rays and how much is dissipated as heat.

Minimum Wavelength of X-rays

X-rays are produced when high-energy electrons strike a metal target in an X-ray tube, and the minimum wavelength of these X-rays is crucial in understanding their penetrating power. The minimum wavelength \( \lambda_{min} \) is determined using the formula \( \lambda_{min} = \frac{hc}{eV} \). Here, \( h \) is the Planck's constant \((6.63 \times 10^{-34} \text{ J.s})\), \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\), \( e \) is the charge of an electron \((1.602 \times 10^{-19} \text{ C})\), and \( V \) is the accelerating voltage.

For an X-ray tube operating at \( 50,000 \text{ V} \), this calculation is as follows:

• \( \lambda_{min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.602 \times 10^{-19} \times 50,000} \approx 0.248 \times 10^{-10} \text{ m} \)

This wavelength value aligns with what is expected from high voltage X-ray tubes, as higher voltage results in shorter, and thus more penetrating, X-ray wavelengths.

Thermal Properties of Materials

When considering materials for an X-ray tube target, their thermal properties are essential due to the significant amount of heat generated. Specifically, it's important to take into account the following aspects:

• **High Melting Temperature**: A suitable target material must have a high melting point. This allows the material to remain stable and not melt under the intense heat produced by the conversion of electric energy into X-rays.
• **Thermal Conductivity**: A high thermal conductivity is preferred over a low one as it allows for faster heat dissipation. Efficient heat dissipation prevents overheating and maintains the integrity of the target material.

In practice, a high melting temperature ensures the longevity of the target material, whereas low thermal conductivity (option (b)) is not ideal, as it results in inefficient heat management. Therefore, when designing or selecting materials for X-ray tubes, both these thermal aspects must be optimized to enhance performance and safety.