Problem 40 - Master Resource Book in JEE Main Physics

Question

The energy flux of sunlight reaching the surface of the earth is \(1.388 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of \(550 \mathrm{~nm} .\) [NCERT] (a) \(3.8 \times 10^{21}\) photon \(/ \mathrm{m}^{2}-\mathrm{s} \quad\) (b) \(4.1 \times 10^{18}\) photon \(/ \mathrm{m}^{2}-5\) (c) \(2.6 \times 10^{19}\) photon \(/ \mathrm{m}^{2}-\mathrm{s}\) (d) \(1.9 \times 10^{20}\) photon \(/ \mathrm{m}^{2}-\mathrm{s}\)

Step-by-Step Solution

Answer

(a) \(3.8 \times 10^{21}\) photons/m²/s

Step 1: Calculate the Energy of One Photon

The energy of a single photon can be calculated using the formula: \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength \(550 \text{ nm} = 550 \times 10^{-9} \text{ m}\). Substituting these values into the equation gives: \[E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}} = 3.614 \times 10^{-19} \text{ J}\]

Step 2: Determine the Number of Photons per Second per Square Meter

To find the number of photons, divide the total energy flux by the energy of one photon. The energy flux provided is \(1.388 \times 10^3 \text{ W/m}^2\), which equals \(1.388 \times 10^3 \text{ J/s/m}^2\). Using the result from Step 1, we calculate: \[ \text{Number of Photons} = \frac{1.388 \times 10^3}{3.614 \times 10^{-19}} = 3.84 \times 10^{21} \text{ photons/m}^2\text{/s}\]

Step 3: Compare with Provided Options

Refer to the options provided in the question: (a) \(3.8 \times 10^{21}\) photon \/\text{m}^{2}-\text{s}, (b) \(4.1 \times 10^{18}\) photon \/\text{m}^{2}-\text{s}, (c) \(2.6 \times 10^{19}\) photon \/\text{m}^{2}-\text{s}, (d) \(1.9 \times 10^{20}\) photon \/\text{m}^{2}-\text{s}. The calculation matches option (a).

Key Concepts

Energy of Photons

Photon energy is a fundamental concept in understanding how light interacts with matter. Each photon carries a specific amount of energy, which can be calculated using the formula:

• \(E = \frac{hc}{\lambda}\)

In this formula:

• \(E\) represents the energy of the photon in Joules.
• \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \text{ J s}\).
• \(c\) is the speed of light in a vacuum, \(3 \times 10^8 \text{ m/s}\).
• \(\lambda\) is the wavelength of the photon in meters.

For our calculation, given the average wavelength of sunlight photons as \(550 \text{ nm}\), converting to meters gives \(550 \times 10^{-9} \text{ m}\). By substituting these values, the energy of a single photon is found to be \(3.614 \times 10^{-19} \text{ J}\). This small fraction of energy, when multiplied by the number of photons hitting a surface, results in the measurable energy we perceive as light.

Wavelength and Frequency

Wavelength and frequency are interconnected properties of light. Wavelength \( (\lambda) \) is the distance between successive crests of a wave, usually measured in meters, while frequency \( (u) \) is the number of wave crests that pass a point in one second, measured in Hertz (Hz). These properties are related by the equation:

• \(c = \lambda u\)

This relationship tells us that wavelength and frequency are inversely proportional; as one increases, the other decreases. When dealing with light, the speed of light \( (c) \) is always \(3 \times 10^8 \text{ m/s}\).Understanding this relationship allows us to transition between the properties to find the energy of photons. For photons in sunlight with a wavelength of \(550 \text{ nm}\), this corresponds to a respective frequency that can be determined using the mentioned formula. Knowing either wavelength or frequency, you can determine other essential characteristics of the photons, aiding in calculating their energy.

Photon Flux Calculations

Photon flux is a measure of the flow of photons, specifically the number of photons passing through a unit area per unit time. In the context of the given exercise, photon flux helps quantify how much light energy in the form of photons reaches the Earth's surface. To determine photon flux:1. **Identify the Energy Flux**: Here, it's given as \(1.388 \times 10^3 \text{ W/m}^2\), which is equivalent to \(1.388 \times 10^3 \text{ J/s/m}^2\).2. **Calculate Photon Energy**: Using the energy of a single photon, which in this example is \(3.614 \times 10^{-19} \text{ J}\).3. **Compute Photon Flux**: Divide the energy flux by the energy of one photon:

• \( \frac{1.388 \times 10^3}{3.614 \times 10^{-19}} = 3.84 \times 10^{21} \text{ photons/m}^2\text{/s}\)

This process yields the number of photons circling a square meter of the Earth's surface every second, offering insight into the intensity of solar radiation received at any moment.