Problem 79 - Master Resource Book in JEE Main Physics

Question

If a source of power \(4 \mathrm{~kW}\) produces \(10^{20}\) photons/sec the radiation belongs to a part of the spectrum called (a) \(\gamma\)-rays (b) X-rays (c) UV-rays (d) microwaves

Step-by-Step Solution

Answer

The radiation belongs to the X-ray part of the spectrum.

Step 1: Understanding Power and Energy

First, we need to remember that power is the rate at which energy is produced. Power in watts (W) is equal to energy per second (Joules per second). In this problem, the power generated is given as \(4\, \text{kW} = 4000\, \text{W}\). Thus, the energy produced per second is \(4000\, \text{J}\).

Step 2: Calculating the Energy Per Photon

Next, we establish the energy per photon. Since we have \(10^{20}\) photons being produced each second, we divide the total energy per second by the number of photons to find the energy per photon: \( \frac{4000\, \text{J}}{10^{20}\, \text{photons}} = 4 \times 10^{-17} \text{J/photon} \).

Step 3: Converting Energy to Wavelength

To find out which part of the electromagnetic spectrum these photons belong to, we need to convert the energy of a photon to a wavelength. This is done using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{J}\cdot\text{s}\), and \( c \) is the speed of light \(3 \times 10^8 \text{m/s}\). Therefore, \( \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4 \times 10^{-17}} \approx 4.97 \times 10^{-11} \text{m} \).

Step 4: Determining the Spectrum Region

Finally, refer to the electromagnetic spectrum where wavelengths are commonly categorized: \(\gamma\)-rays (less than \(10^{-12}\) m), X-rays (\(10^{-12}\) to \(10^{-9}\) m), UV-rays (\(10^{-9}\) to \(10^{-7}\) m), and microwaves (\(10^{-3}\) to \(10^{-1}\) m). With the wavelength calculation of \(4.97 \times 10^{-11} \text{m}\), these photons fit into the X-ray category.

Key Concepts

Photon Energy

In the world of electromagnetic waves, understanding photon energy is crucial. Photons are elementary particles that carry energy, and their energy is directly related to their frequency and inversely related to their wavelength. The formula that connects these concepts is given by \( E = h \times f \) or, as used in wavelength calculations, \( E = \frac{hc}{\lambda} \). Here:

• \( E \) signifies energy in joules.
• \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{J}\cdot\text{s} \).
• \( f \) represents frequency.
• \( \lambda \) (lambda) is the wavelength.
• \( c \) indicates the speed of light in vacuum, approximately \( 3 \times 10^8 \text{m/s} \).

In practice, photon energy is crucial for identifying the part of the electromagnetic spectrum a given radiation belongs to. Each category—from microwaves to gamma rays—has a characteristic energy range that helps scientists determine the properties of the radiation.

X-rays

X-rays are a fascinating component of the electromagnetic spectrum. They have wavelengths ranging from about \( 10^{-12} \text{m} \) to \( 10^{-9} \text{m} \), and this precise wavelength categorization is essential.

What sets X-rays apart from other types of radiation is their ability to penetrate various substances. This property makes them useful in numerous applications:

• Medical imaging: X-rays are extensively used in healthcare to view the inside of the body.
• Security: They help in scanning luggage and cargo in transportation sectors.
• Scientific research: In laboratories, X-rays are used for crystallography to determine structures of molecules.

The energy of X-ray photons is significantly higher than that of visible light or UV rays, which is why they can go through materials opaque to lower energy radiation. However, this also means that X-ray radiation exposure must be controlled to avoid tissue damage.

Wavelength Calculation

To determine where radiation falls within the electromagnetic spectrum, calculating the wavelength from energy is a pivotal step. Using the relationship \( E = \frac{hc}{\lambda} \), we can rearrange to find:

• The wavelength \( \lambda = \frac{hc}{E} \).

In the given exercise, with the photon energy calculated as \( 4 \times 10^{-17} \text{J} \), this formula allows us to compute the wavelength:\[\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-17}} \approx 4.97 \times 10^{-11} \text{m}\]Understanding the wavelength derived from energy calculations is not just about number crunching. It is a critical step in identifying which section of the electromagnetic spectrum the photons belong to.

This allows scientists to determine the suitability of the radiation for various practical applications, such as medical imaging with X-rays or other advanced material analysis techniques.

Electromagnetic Spectrum Categories

The electromagnetic spectrum holds the entire range of electromagnetic radiation, from low-frequency radio waves to high-energy gamma rays. Each category on the spectrum is defined by a specific wavelength range. Understanding each category is vital for classifying radiation correctly.

Here's a quick rundown:

• **Radio Waves:** Longest wavelengths, greater than \( 10^{-1} \text{m} \). Used in communication.
• **Microwaves:** Ranging from \( 10^{-3} \text{m} \) to \( 10^{-1} \text{m} \). Handy for cooking and some communications.
• **Infrared:** Lies between \( 10^{-6} \text{m} \) and \( 10^{-3} \text{m} \). It's the heat we feel from the sun.
• **Visible Light:** Approximately \( 10^{-7} \text{m} \) to \( 10^{-6} \text{m} \), the only part of the spectrum visible to the human eye.
• **Ultraviolet (UV):** Ranges from \( 10^{-9} \text{m} \) to \( 10^{-7} \text{m} \). Can cause skin tanning or burns.
• **X-rays:** Aa bit shorter than UV, from \( 10^{-12} \text{m} \) to \( 10^{-9} \text{m} \). Great for imaging inside the body.
• **Gamma Rays:** Shortest wavelengths, less than \( 10^{-12} \text{m} \). Associated with nuclear reactions.

Each category of the electromagnetic spectrum opens up a window to different practical applications and scientific insights, enabling technology to cater to various needs from simple communications to complex medical procedures.