Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

Electron energy of a photon is given as: \(\Delta \mathrm{E} /\) atom \(=3.03 \times 10^{-19} \mathrm{~J}\) atom \(^{-1}\) then, the wavelength of the photon is (a) \(6.56 \mathrm{~nm}\) (b) \(65.6 \mathrm{~nm}\) (c) \(656 \mathrm{~nm}\) (d) \(0.656 \mathrm{~nm}\) Given, h (Planck constant) \(=6.63 \times 10^{-34} \mathrm{Js} \mathrm{c}\) (velocity of light \()=3.00 \times 10^{8} \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The wavelength of the photon is 656 nm.

Step 1: Understand the Problem

We need to find the wavelength of a photon given its energy, \(\Delta \mathrm{E}\) atom = \(3.03 \times 10^{-19} \mathrm{~J}\) atom \(^{-1}\). We will use these constants: Planck's constant \(h = 6.63 \times 10^{-34} \mathrm{Js}\) and speed of light \(c = 3.00 \times 10^{8} \mathrm{~ms}^{-1}\).

Step 2: Use Energy-Wavelength Relationship

We use the energy-wavelength relation: \[ E = \frac{hc}{\lambda} \]where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. We need to solve for \(\lambda\).

Step 3: Solve for Wavelength

Rearrange the equation from Step 2:\[ \lambda = \frac{hc}{E} \]Substituting the given values into the equation, we get:\[ \lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.03 \times 10^{-19}} \]

Step 4: Calculate Wavelength

Calculate the value:\[ \lambda = \frac{1.989 \times 10^{-25}}{3.03 \times 10^{-19}} \]\[ \lambda \approx 656 \times 10^{-9} \] m or \(656\) nm.

Key Concepts

Energy-Wavelength Relation

The energy-wavelength relation is a cornerstone of quantum physics. It connects the energy of a photon to its wavelength using the formula:

• \[ E = \frac{hc}{\lambda} \]

Here, \( E \) represents the energy of the photon, \( h \) is the Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.
This relationship helps us understand how electromagnetic radiation behaves. When the energy \( E \) of the photon increases, the wavelength \( \lambda \) decreases, and vice-versa.
In the context of this exercise, we are given the energy of the photon and need to find its wavelength. By rearranging the formula to solve for \( \lambda \), we get:

• \[ \lambda = \frac{hc}{E} \]

This shows how we can use known values to calculate the unknown wavelength of the photon.

Planck's Constant

Planck's constant is a fundamental quantity in physics. It is symbolized by \( h \), and has a value of approximately \( 6.63 \times 10^{-34} \) Js. This tiny constant plays a massive role in quantum mechanics. It acts as the proportionality factor between the energy \( E \) of a photon and the frequency \( u \) of its corresponding electromagnetic wave. The relation is given by:

• \[ E = hu \]

This constant not only helps define the energy levels in atoms but also underpins the energy-wavelength relation where \( h \) helps connect and balance the units involved in the physical quantities. It provides the basis for numerous equations in quantum physics including the energy-wavelength relation,which is particularly useful in understanding atomic scale phenomena.

Photon Energy

Photon energy is the energy carried by a single photon. It's a fundamental concept when discussing electromagnetic waves, light, and their interactions with matter. The energy of a photon is directly proportional to its frequency \( u \) with the relationship:

• \[ E = hu \]

In the exercise given, you can also see photon energy linked to its wavelength by this formula rearranged from the energy-wavelength relation:

• \[ E = \frac{hc}{\lambda} \]

In practical terms, photon energy determines how photons interact with materials. Higher energy photons, such as X-rays and gamma rays, have enough energy to ionize atoms, while lower energy photons, such as visible or infrared light, do not. It's essential to calculate photon energy accurately when studying the effects of electromagnetic radiation in various physics and engineering applications.