Problem 82 - Master Resource Book in JEE Main Physics

Question

An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of \(45 \times 10^{-2} \AA\). The maximum energy of a photon in the radiation in \(\mathrm{eV}\) is \(\left(h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}, c=3 \times 10^{8} \mathrm{~ms}^{-1}\right)\) (a) 27,500 (b) 22,500 (c) 17,500 (d) 12,500

Step-by-Step Solution

Answer

The maximum energy of a photon is approximately 27,500 eV, so the answer is (a) 27,500.

Step 1: Identify the Given Values and Required Formula

We are given the shortest wavelength \( \lambda = 45 \times 10^{-2} \AA \). Convert this to meters: \( 1 \AA = 10^{-10} \text{ m} \), so \( \lambda = 45 \times 10^{-12} \text{ m} \). We need to find the maximum energy of a photon using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.62 \times 10^{-34} \text{ J-s} \) and \( c \) is the speed of light \( 3 \times 10^{8} \text{ m/s} \).

Step 2: Calculate the Energy in Joules

Substitute the given values into the formula:\[E = \frac{(6.62 \times 10^{-34} \text{ J-s}) \times (3 \times 10^{8} \text{ m/s})}{45 \times 10^{-12} \text{ m}}\]Calculate the numerator: \( 6.62 \times 3 = 19.86 \times 10^{-34+8} = 19.86 \times 10^{-26} \). Divide by \( 45 \times 10^{-12} \) to get:\[E \approx \frac{19.86 \times 10^{-26}}{45 \times 10^{-12}} = 4.41 \times 10^{-15} \text{ J}\]

Step 3: Convert Energy from Joules to Electron Volts (eV)

To convert the energy from joules to electron volts, use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):\[E_{eV} = \frac{4.41 \times 10^{-15} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}}\]Calculate the result:\[E_{eV} \approx 27,562.5 \, \text{eV}\]

Step 4: Match the Calculated Energy to the Closest Answer

The calculated energy in electron volts is approximately 27,562.5 eV. Comparing this value to the given options, 27,500 eV is the closest match. Therefore, the correct answer is (a) 27,500 eV.

Key Concepts

Photon Energy Calculation

To calculate the energy of a photon, we use the formula: \[ E = \frac{hc}{\lambda} \] where:

• \( E \) is the energy of the photon in joules (J).
• \( h \) is Planck's constant, which is \( 6.62 \times 10^{-34} \text{ J-s} \).
• \( c \) is the speed of light, \( 3 \times 10^{8} \text{ m/s} \).
• \( \lambda \) is the wavelength of the photon.

This formula shows the relationship between the size of a photon's wavelength and its energy. The shorter the wavelength, the higher the energy of the photon. This is particularly important in the study of X-rays due to their short wavelengths and high energies.

Wavelength Conversion

Wavelength often needs to be converted into different units to be used in formulas. In this case, the wavelength is given in angstroms (Å), a common unit in physics. To convert to meters (the standard unit for scientific calculations), use the conversion factor:

• \(1 \text{ Å} = 10^{-10} \text{ meters} \)

For example, a wavelength of \( 45 \times 10^{-2} \text{ Å} \) converts to meters by multiplying by \( 10^{-10} \), resulting in \( 45 \times 10^{-12} \text{ meters} \). Using consistent units is crucial for ensuring accurate calculations.

Planck's Constant

Planck's constant \( h \) is a fundamental constant in physics, playing a vital role in the field of quantum mechanics. Its value is \( 6.62 \times 10^{-34} \text{ J-s} \). Here's what makes Planck's constant special:

• It relates the energy of a photon to its frequency, which is essential for understanding phenomena like the photoelectric effect.
• Used widely in equations that describe the particle-like properties of light.

It acts as a bridge between the wave theories of light and the emergent quantum properties.

Speed of Light

The speed of light, represented as \( c \), is a universal physical constant important in many areas of physics. Its value is \( 3 \times 10^{8} \text{ m/s} \). What you need to know about the speed of light:

• It is the fastest speed at which information or matter can travel through space.
• Used in calculations to relate energy, frequency, and wavelength.
• Plays a key role in equations governing relativity and electromagnetism.

Understanding its role in these calculations is fundamental to topics ranging from optics to quantum mechanics.

Electron Volt Conversion

The electron volt (eV) is a unit of energy often used in atomic and particle physics. To convert energy from joules to electron volts, we use a specific conversion factor. 1 electron volt is equivalent to \( 1.6 \times 10^{-19} \text{ J} \). When you have energy in joules and need to convert to eV, divide by \( 1.6 \times 10^{-19} \). For example, with \( 4.41 \times 10^{-15} \text{ J} \), the conversion to eV is: \[ E_{eV} = \frac{4.41 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 27,562.5 \text{ eV} \] Using electron volts allows scientists to work with more manageable numbers when dealing with tiny energy levels in subatomic processes.