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, option (c).

Step 1: Understand the Formula Involved

We're given the energy of a photon and need to find its wavelength. The energy of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \]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 find.

Step 2: Rearrange to Solve for Wavelength

To find the wavelength \( \lambda \), we need to rearrange the formula:\[ \lambda = \frac{hc}{E} \]

Step 3: Insert Given Values into Formula

Substitute the given values into the equation:- \( E = 3.03 \times 10^{-19} \) J- \( h = 6.63 \times 10^{-34} \) Js- \( c = 3.00 \times 10^{8} \) m/sThe formula becomes:\[ \lambda = \frac{(6.63 \times 10^{-34} \text{ Js})(3.00 \times 10^{8} \text{ m/s})}{3.03 \times 10^{-19} \text{ J}} \]

Step 4: Perform the Calculation

Calculate the above expression:\[ \lambda = \frac{1.989 \times 10^{-25} \text{ Js m/s}}{3.03 \times 10^{-19} \text{ J}} = 6.56 \times 10^{-7} \text{ m} \]

Step 5: Convert Wavelength to Nanometers

The wavelength in meters can be converted to nanometers by multiplying with \(10^9\):\[ \lambda = 6.56 \times 10^{-7} \text{ m} \times 10^9 \frac{\text{nm}}{\text{m}} = 656 \text{ nm} \]

Step 6: Choose the Correct Option

Compare the calculated wavelength (656 nm) to the provided options. Option (c) is 656 nm, which matches our result.

Key Concepts

Planck's Constant

Planck's constant is a fundamental constant in physics, symbolized as \( h \), which is crucial in the study of quantum mechanics.
It connects the energy of a photon with the frequency of its electromagnetic wave.
This constant has a value of \( 6.63 \times 10^{-34} \) Joules per second (Js). It's a small number because the quantized nature of energy at atomic scales is very tiny.

Understanding Planck's constant allows us to comprehend how energy is quantized. It signifies that energy levels are not continuous, but rather occur in discrete packets called "quanta."

• It supports the concept that light exhibits both wave-like and particle-like properties.
• Forms the basis for the equation \( E = hf \), where \( E \) is energy, \( h \) is Planck's constant, and \( f \) is frequency.

Knowing \( h \) is essential for calculations involving energy and wavelength, as seen in the formula \( E = \frac{hc}{\lambda} \).
Be sure to remember its small magnitude when dealing with problems in quantum mechanics.

Speed of Light

The speed of light, denoted as \( c \), is an important constant representing how fast light travels in a vacuum.
It has a value of \( 3.00 \times 10^8 \) meters per second (m/s).
The speed of light is one of the most crucial and invariant speeds in nature, underpinning all electromagnetic phenomena.

This constant allows us to understand how quickly information or energy can be transmitted through space.

• The constancy of light speed supports Einstein's theory of relativity.
• It's vital for calculating other properties of light, such as wavelength and energy.

The relationship between wavelength and frequency is linked by the speed of light with the equation \( c = \lambda f \), where \( \lambda \) is the wavelength and \( f \) is the frequency.
This formula helps us to convert between these properties, ensuring precise calculations in topics like photon energy and wavelengths.

Wavelength Calculation

In physics, calculating the wavelength of a photon when its energy is known involves using the relationship between energy, Planck's constant, and the speed of light.
The key formula is \( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.
This equation can be rearranged to solve for the wavelength as \( \lambda = \frac{hc}{E} \).

When dealing with photon energy problems, follow these steps:

• Firstly, recognize the values you're working with: Planck's constant \( h = 6.63 \times 10^{-34} \) Js, speed of light \( c = 3.00 \times 10^8 \) m/s, and the given energy \( E \).
• Substitute these values into the rearranged formula \( \lambda = \frac{hc}{E} \).
• Complete the calculations to find \( \lambda \), ensuring units are consistent for accurate results.

Once you have \( \lambda \) in meters, you may need to convert it to more practical units like nanometers by multiplying by \( 10^9 \).
This conversion allows you to compare your answer with standard options, often necessary in exercises and problem-solving situations.