Problem 42 - Master Resource Book in JEE Main Physics

Question

For light of wavelength \(5000 \AA\), photon energy is nearly \(2.5 \mathrm{eV}\). For \(X\)-rays of wavelength \(1 \AA\), the photon energy will be close to [a) \([2.5 \div 5000] \mathrm{eV}\) (b) \(\left[2.5 \div(5000)^{2}\right] \mathrm{eV}\) (c) \([2.5 \times 5000] \mathrm{eV}\) (d) \(\left[2.5 \times(5000)^{2}\right] \mathrm{eV}\)

Step-by-Step Solution

Answer

The photon energy for 1 Å X-rays is 12500 eV, so option (c) is correct.

Step 1: Understanding Photon Energy

The energy of a photon is given by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. The shorter the wavelength, the higher the energy of the photon.

Step 2: Determining the Proportionality

The energy of a photon is inversely proportional to its wavelength, \( E \propto \frac{1}{\lambda} \). Therefore, if the wavelength decreases, the energy increases proportionally.

Step 3: Relating Photon Energies for Different Wavelengths

Let \( E_1 \) and \( E_2 \) be the energies of photons with wavelengths \( \lambda_1 \) and \( \lambda_2 \) respectively. According to the formula, \( \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} \).

Step 4: Applying Wavelengths to the Formula

Given \( \lambda_1 = 5000 \AA \) and \( E_1 = 2.5 \, \mathrm{eV} \). For the \( X \)-rays, \( \lambda_2 = 1 \AA \). Then, \( \frac{E_2}{2.5} = \frac{5000}{1} \), so \( E_2 = 2.5 \times 5000 \).

Step 5: Calculating Photon Energy for X-rays

Substitute the values: \( E_2 = 2.5 \times 5000 = 12500 \, \mathrm{eV} \). Thus, the correct answer is option (c) \([2.5 \times 5000] \, \mathrm{eV}\).

Key Concepts

Wavelength

When we talk about the wavelength of a wave, including light or other electromagnetic radiation such as X-rays, we refer to the distance between two consecutive peaks or troughs in the wave. Wavelength is denoted by the Greek letter lambda (\( \lambda \)) and is typically measured in meters or angstroms, where 1 angstrom (\( 1 \AA \)) is equal to \( 10^{-10} \) meters.

Wavelength is a critical factor in determining the properties of electromagnetic radiation. It determines not only the type of radiation but also its energy. Shorter wavelengths correspond to higher energy levels, while longer wavelengths are associated with lower energy. This is due to the inverse relationship between energy and wavelength, which means as one increases, the other decreases.

Understanding how wavelength affects energy is central to interpreting many physical phenomena, from visible light to the powerful X-rays used in medical imaging.

X-rays

X-rays are a form of electromagnetic radiation, like visible light, but they possess much shorter wavelengths and consequently much higher energies. These wavelengths typically range from about 0.01 to 10 nanometers or equivalently, 0.1 to 100 angstroms.

Because of their high energy, X-rays can penetrate various materials, which is why they are widely used in medical imaging to view the inside of human bodies, particularly bones. Additionally, their properties make them valuable in various industrial applications such as inspecting welding seams and materials testing.

The shorter wavelength and higher energy of X-rays compared to visible light make them invaluable tools across many fields. However, it's important to handle X-rays with care due to their ability to cause damage to living tissues upon prolonged exposure.

Planck's constant

Planck's constant is a fundamental quantity in quantum mechanics, denoted by \( h \). It describes the quantization of energy levels in quantum systems, implying that energy is not continuous but rather comes in discrete "chunks" called quanta.

The value of Planck's constant is approximately \( 6.626 \times 10^{-34} \) Joule second (Js). This constant is crucial in many areas of physics, especially in calculations that involve energy and frequency, such as determining the energy of a photon.

When dealing with the energy of photons, the equation \( E = \frac{hc}{\lambda} \) links these concepts together, where \( c \) is the speed of light. This highlights how fundamental Planck's constant is in understanding the relationship between energy and frequency or wavelength, facilitating insights into the behaviors of particles at microscopic scales.