Problem 8 - Master Resource Book in JEE Main Physics

Question

There are two sources of light each emitting with a power of \(100 \mathrm{~W}\). One emits X-rays of wavelength \(1 \mathrm{~nm}\) and the other visible light of wavelength \(500 \mathrm{~nm} .\) Find the ratio of number of photons of X-rays in the photons of visible light of the given wavelength? (a) \(1: 500\) (b) \(1: 250\) (c) \(1: 20\) (d) 100

Step-by-Step Solution

Answer

Ratio of photons: 1:500 (Answer a).

Step 1: Understanding the Relationship between Energy and Wavelength

The energy of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \]where:- \(h\) is Planck's constant approximately \(6.626 \times 10^{-34} \text{ Js}\).- \(c\) is the speed of light, approximately \(3 \times 10^8 \text{ m/s}\).- \(\lambda\) is the wavelength in meters.

Step 2: Calculate the Energy of X-ray Photons

For X-rays with a wavelength of \(1 \text{ nm}\):Convert the wavelength to meters: \(\lambda_{\text{X-ray}} = 1 \times 10^{-9} \text{ m}\).Substitute into the energy formula: \[ E_{\text{X-ray}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1 \times 10^{-9}} \]\[ E_{\text{X-ray}} = 1.9878 \times 10^{-16} \text{ J} \]

Step 3: Calculate the Energy of Visible Light Photons

For visible light with a wavelength of \(500 \text{ nm}\):Convert the wavelength to meters: \(\lambda_{\text{vis}} = 500 \times 10^{-9} \text{ m}\).Substitute into the energy formula: \[ E_{\text{vis}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} \]\[ E_{\text{vis}} = 3.9756 \times 10^{-19} \text{ J} \]

Step 4: Calculate the Number of Photons from Each Source

The total power of each source is \(100 \text{ W}\), which means \(100 \text{ J/s}\). The number of photons emitted per second can be determined by dividing the power by the energy of a single photon for each source:For X-rays: \[ N_{\text{X-ray}} = \frac{100}{1.9878 \times 10^{-16}} \approx 5.03 \times 10^{17} \text{ photons/s} \]For visible light: \[ N_{\text{vis}} = \frac{100}{3.9756 \times 10^{-19}} \approx 2.52 \times 10^{20} \text{ photons/s} \]

Step 5: Determine the Ratio of Photons

To find the ratio of the number of X-ray photons to the number of visible light photons:\[ \text{Ratio} = \frac{N_{\text{X-ray}}}{N_{\text{vis}}} = \frac{5.03 \times 10^{17}}{2.52 \times 10^{20}} \approx 0.002 \]This ratio simplifies approximately to:\[ 1: 500 \]

Key Concepts

X-ray Photons

X-ray photons are a type of electromagnetic radiation with wavelengths ranging from 0.01 to 10 nanometers. They are part of the electromagnetic spectrum and sit between ultraviolet light and gamma rays. X-rays are typically used in medical imaging to create images of the interior of the body, such as bones and organs, by passing through lighter materials but being absorbed by denser substances like bones. Understanding X-ray photons involves knowing that they have high energy due to their short wavelength. The energy of a photon increases as the wavelength decreases. This is expressed in the energy formula \( E = \frac{hc}{\lambda} \) . For instance, an X-ray photon with a wavelength of \(1 \text{ nm}\) possesses significantly more energy than many other forms of light.Knowing this characteristic helps us comprehend why X-ray photons are more powerful and can penetrate materials to reveal internal structures in medical diagnostics.

Visible Light Photons

Visible light photons are photons within the part of the electromagnetic spectrum humans can see. This spectrum includes wavelengths from approximately 380 to 750 nanometers. These photons are responsible for illuminating the world around us, giving us color and enabling vision. They are much less energetic compared to X-ray photons due to their longer wavelengths. For visible light with a wavelength of \(500 \text{ nm}\), the energy per photon is calculated using the formula \( E = \frac{hc}{\lambda} \). This formula shows that the longer the wavelength (like visible light), the less energy a photon has. Visible light photons are crucial for life's everyday processes; from enabling photosynthesis in plants to influencing human circadian rhythms by signaling daylight. Their energy levels make them safe for direct human exposure, unlike their X-ray counterparts.

Planck's Constant

Planck's constant is a fundamental constant in physics, denoted by \( h \), with a value of approximately \(6.626 \times 10^{-34} \text{ Js}\). It is a crucial factor in the formula used to determine the energy of photons: \( E = \frac{hc}{\lambda} \), where \( c \) is the speed of light (approximately \(3 \times 10^8 \text{ m/s}\)). Planck's constant is named after Max Planck, who was a pioneer in quantum theory. It is central to understanding the behavior of particles at a quantum level, linking the frequency of a photon to its energy. Without Planck's constant, predicting and measuring the energy of photons across various wavelengths would be impossible. In practical terms, it helps us understand not only how different photons of light carry energy, as seen when comparing X-ray and visible light photons, but also the broader mechanics of atomic and subatomic processes.

Wavelength Conversion

Wavelength conversion is a necessary step when working with photon energy calculations. Photons travel with different wavelengths depending on their position in the electromagnetic spectrum. To accurately calculate the energy of photons, their wavelength often needs to be converted into meters. For example, X-ray photons with a wavelength of \(1 \text{ nm}\) are converted to \(1 \times 10^{-9} \text{ m}\) for precise computation. Similarly, visible light photons with a wavelength of \(500 \text{ nm}\) need to be converted to \(500 \times 10^{-9} \text{ m}\) . Wavelength conversion ensures that the calculations remain consistent with standard units used in scientific equations. This conversion underpins the understanding of energy differences between various forms of light and helps perform accurate comparisons, as seen in determining the ratio of photons between different light sources.