Problem 5 - Master Resource Book in JEE Main Physics

Question

An X-ray tube with copper target emit \(K_{\alpha}\) line of wavelength \(1.50 \AA\). What should be the minimum voltage through which electrons one to be accelerated to produce this wavelength of X-rays. \(\left(h=6.6 \times 10^{-34} \mathrm{Js}, c=3 \times 10^{8} \mathrm{~ms}^{-1}\right)\) (a) \(82.8 \mathrm{~V}\) (b) \(8280 \mathrm{~V}\) (c) \(82801 \mathrm{~V}\) (d) \(828 \mathrm{~V}\)

Step-by-Step Solution

Answer

The minimum voltage required is 8280 V.

Step 1: Understanding the Problem

The problem asks us to find the minimum voltage required to accelerate electrons to produce X-ray of given wavelength. We will use the relationship between energy, wavelength, and voltage.

Step 2: Use Energy-Wavelength Formula

The energy of a photon is given by the equation:\[ E = \frac{hc}{\lambda} \]where \( h \) is Planck's constant \( (6.6 \times 10^{-34} \ \mathrm{Js}) \), \( c \) is the speed of light \( (3 \times 10^8 \ \mathrm{ms}^{-1}) \), and \( \lambda \) is the wavelength. Substituting the given wavelength \( 1.50 \ \mathrm{\AA} = 1.50 \times 10^{-10} \ \mathrm{m} \):\[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.50 \times 10^{-10}} \]

Step 3: Calculate the Energy

Compute the expression for energy:\[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.50 \times 10^{-10}} = 1.32 \times 10^{-15} \ \mathrm{J} \]

Step 4: Convert Energy to Electron Volts

To find the voltage, convert the energy in joules to electron volts, knowing that \( 1 \, \mathrm{eV} = 1.6 \times 10^{-19} \ \mathrm{J} \). Use:\[ E = \frac{1.32 \times 10^{-15}}{1.6 \times 10^{-19}} \ \mathrm{eV} \]

Step 5: Compute the Voltage

Calculate the energy in electron volts:\[ E = \frac{1.32 \times 10^{-15}}{1.6 \times 10^{-19}} = 8250 \ \mathrm{eV} \]Thus, the electrons must be accelerated through a potential difference of approximately 8280 V.

Key Concepts

Energy-Wavelength Relationship

When discussing X-ray production, it's crucial to understand the connection between a photon's energy and its wavelength. Photons, the basic units of light, have energy that is inversely related to their wavelength. The shorter the wavelength, the higher the energy. This relationship can be expressed by the formula: \[ E = \frac{hc}{\lambda} \]where \( E \) is the photon's energy, \( h \) is Planck’s constant (\(6.6 \times 10^{-34} \ \mathrm{Js}\)), \( c \) is the speed of light (\(3 \times 10^8 \ \mathrm{m/s}\)), and \( \lambda \) is the wavelength. In the given exercise, the X-ray wavelength is \(1.50 \ \mathrm{\AA}\), which is equivalent to \(1.50 \times 10^{-10} \ \mathrm{m}\).

• As the wavelength decreases, energy increases.
• For X-rays, which have very short wavelengths, the photon's energy is relatively high.

This formula is the starting point for calculating the energy required to produce a specific type of X-ray photon.

Electron Acceleration

To produce X-rays, electrons are accelerated to high speeds and then suddenly decelerated when they hit a metal target. The energy these electrons gain from the acceleration is what eventually transforms into X-ray photons. The speed and energy of these electrons are determined by the voltage applied across the X-ray tube. The relationship between the voltage \( V \), charge of an electron \( e \), and the energy it gains is given by:\[ E = eV \]where \( e \) is the elementary charge \((1.6 \times 10^{-19} \ \mathrm{C})\). The energy gained by an electron when accelerated over a potential difference \( V \) is directly used to convert into the energy of an X-ray photon.

The higher the voltage, the more energy the electron gains.

Higher energy electrons produce higher energy, shorter wavelength X-rays.

Photon Energy Calculation

Calculating the energy of a photon helps determine the minimum energy that electrons need to possess to create a specific X-ray wavelength. As per the exercise, we calculated the energy using the energy-wavelength relationship:\[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.50 \times 10^{-10}} \]This calculation resulted in:\[ E = 1.32 \times 10^{-15} \ \mathrm{J} \]

• This energy value in joules is the absolute minimum energy required for an electron to produce the given X-ray wavelength.
• Once obtained, you can convert this energy to different units for easier understanding and application, such as into electron volts.

Electron Volt Conversion

After calculating the energy of a photon in joules, it's often useful to convert this energy into electron volts \((\mathrm{eV})\), a more practical and common unit used in atomic and particle physics. The conversion from joules to electron volts utilizes the fact that:\[ 1 \ \mathrm{eV} = 1.6 \times 10^{-19} \ \mathrm{J} \]To convert the calculated photon energy:\ \[ E = \frac{1.32 \times 10^{-15}}{1.6 \times 10^{-19}} \ \mathrm{eV} \]This gives:\[ E \approx 8250 \ \mathrm{eV} \]

• Electron volts simplify complex calculations and provide a more intuitive understanding of energy scales, especially in atomic-scale phenomena.
• In this exercise, the result indicates the voltage needed to impart the calculated energy to an electron, thus allowing it to produce the specific X-ray wavelength.