Problem 46 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Wavelength of photon which have energy equal to average of energy of photons with \(\lambda_{1}=4000 \AA\) and \(\lambda_{2}=6000 \AA\) will be (a) \(5000 \AA\) (b) \(4800 \AA\) (c) \(9600 \AA\) (d) \(2400 \AA\)

Step-by-Step Solution

Answer

The wavelength of a photon with the average energy of photons with wavelengths \(4000 \AA\) and \(6000 \AA\) is \(4800 \AA\).

Step 1: Calculate the Energies of Given Photons

Using the formula for the energy of a photon, E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon, compute the energy for both wavelengths: For \(\lambda_{1} = 4000 \AA\): \(E_{1} = \frac{hc}{\lambda_{1}}\)For \(\lambda_{2} = 6000 \AA\): \(E_{2} = \frac{hc}{\lambda_{2}}\)Note that \(hc\) is a constant, so it will be the same for both equations.

Step 2: Find the Average Energy

Find the average energy \(\bar{E}\) of the two photons by calculating the average of \(E_{1}\) and \(E_{2}\): \(\bar{E} = \frac{E_{1} + E_{2}}{2} = \frac{hc}{2}\left(\frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}\right)\).

Step 3: Calculate the Wavelength for the Average Energy

Using the average energy \(\bar{E}\) to find the wavelength \(\lambda_{avg}\) of a photon with this energy: \(\bar{E} = \frac{hc}{\lambda_{avg}}\).Then solve for \(\lambda_{avg}\): \(\lambda_{avg} = \frac{hc}{\bar{E}}\).Since \(hc\) is a constant, it can be removed from the equation while comparing relative wavelengths, which simplifies to: \(\lambda_{avg} = \frac{2\lambda_{1}\lambda_{2}}{\lambda_{1} + \lambda_{2}}\).

Step 4: Substitute the Values to Find \(\lambda_{avg}\)

Put \(\lambda_{1} = 4000 \AA\) and \(\lambda_{2} = 6000 \AA\) into the equation from step 3 to get \(\lambda_{avg}\):\[\lambda_{avg} = \frac{2 \times 4000 \times 6000}{4000 + 6000} = \frac{48000000}{10000} = 4800 \AA\].

Key Concepts

Planck's Constant

Planck's constant, denoted by the symbol \(h\), is a fundamental constant that plays a vital role in quantum mechanics. It relates the energy of a photon, which is the smallest discrete amount or quantum of electromagnetic radiation, to its frequency. The value of Planck's constant is approximately \(6.626 \times 10^{-34} \text{Js}\t (joule seconds).\)

Understanding Planck's constant is crucial when dealing with photon energy calculations because it acts as a bridge between the macroscopic and quantum worlds. It's essential to comprehend how this constant leads us from a wavelength of a photon to its energy. In simple terms, through Planck's constant, we can see that energy is quantized, which means it exists in individual units rather than on a continuous spectrum.

Speed of Light

The speed of light in a vacuum, represented by the symbol \(c\), is another fundamental physical constant important for understanding the behavior of electromagnetic radiation, including photons. The widely accepted value of the speed of light is approximately \(3 \times 10^8 \text{m/s}\t (meters per second).\)

When calculating the energy of a photon, the speed of light combines with Planck's constant to determine how energy is distributed across varying wavelengths. This helps us appreciate that light is not only a wave, but also a particle, and that it travels at a constant speed irrespective of the energy that each photon carries.

Wavelength and Energy Relationship

The relationship between wavelength and energy for a photon is inversely proportional. This means that longer wavelengths correspond to lower energy photons, and shorter wavelengths to higher energy photons. This relationship is expressed in the formula \(E = \frac{hc}{\t\text{λ}}\), where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \t\text{λ}\t is the wavelength of the photon.

Understanding this equation is fundamental when calculating the energy of a photon based on its wavelength. For instance, visible light has wavelengths in the range of about 400 nm to 700 nm and includes a variety of energies within that spectrum. The colors we see are a direct result of these differences in photon energies.

Average Energy of Photons

When dealing with multiple photons, it's often necessary to find the average energy. The average energy of photons can be understood as the arithmetic mean of individual photon energies. In a conceptual exercise, if two photons have energies \(E_1\) and \(E_2\), the average energy is calculated as \(\bar{E} = \frac{E_1 + E_2}{2} \).

This concept is useful when considering how light of various wavelengths can mix to yield an effective average wavelength, as seen in problem-solving. This average effectively represents the combined effect of multiple photons acting together, an important consideration in wave optics and quantum mechanics.