Problem 13 - Master Resource Book in JEE Main Physics

Question

If the energy of photons corresponding to the wavelength of \(6000 \mathrm{~A}\) is \(3.2 \times 10^{-19} \mathrm{~J}\), the photon energy for a wavelength of 4000 A will be (a) \(1.11 \times 10^{-19} \mathrm{~J}\) (b) \(2.22 \times 10^{-19} \mathrm{~J}\) (c) \(4.40 \times 10^{-19} \mathrm{~J}\) (d) \(4.80 \times 10^{-19} \mathrm{~J}\)

Step-by-Step Solution

Answer

The photon energy at 4000 Å is (d) \( 4.80 \times 10^{-19} \, \text{J} \).

Step 1: Understand the Relationship between Wavelength and Energy

The energy of a photon is inversely proportional to its wavelength. This relationship is expressed by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of the photon, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

Step 2: Compare Energy at Different Wavelengths

Given two wavelengths \( \lambda_1 = 6000 \) Å and \( \lambda_2 = 4000 \) Å. We can use the formula \( \frac{E_2}{E_1} = \frac{\lambda_1}{\lambda_2} \) to find the energy \( E_2 \) at wavelength \( \lambda_2 \).

Step 3: Calculate the Energy Ratio

Substitute \( \lambda_1 = 6000 \, \text{Å} \) and \( \lambda_2 = 4000 \, \text{Å} \) into the equation: \( \frac{E_2}{E_1} = \frac{6000}{4000} = 1.5 \). Thus, \( E_2 = 1.5 \times E_1 \).

Step 4: Compute the Energy at 4000 Å

Substitute the given energy at 6000 Å, \( E_1 = 3.2 \times 10^{-19} \, \text{J} \), into the equation: \( E_2 = 1.5 \times 3.2 \times 10^{-19} \). Calculate \( E_2 = 4.8 \times 10^{-19} \, \text{J} \).

Step 5: Determine the Correct Answer

Compare the calculated energy \( 4.8 \times 10^{-19} \, \text{J} \) with the options provided. The correct answer is (d) \( 4.80 \times 10^{-19} \, \text{J} \).

Key Concepts

Wavelength and Energy Relationship

The connection between wavelength and energy is fundamental in understanding how photons behave. To put it simply, photons are particles of light, and their energy is inversely proportional to their wavelength. This means that when the wavelength of a light wave decreases, the energy of its photons increases, and vice versa. So, shorter wavelengths correspond to higher energy photons and longer wavelengths to lower energy photons.

This relationship is mathematically expressed by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of a photon, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. It's essential to remember:

• Short wavelength = high energy
• Long wavelength = low energy

Understanding this inverse relationship helps in predicting how changes in wavelength affect the energy of light, crucial for applications like determining photon energies in different light spectrums.

Planck's Constant

Planck's constant is a key figure in quantum mechanics, representing the smallest action in the physical world. Its value is fundamental to many calculations involving photons and their energies.

Denoted as \( h \), Planck's constant has a value of approximately \( 6.626 \times 10^{-34} \) Joule seconds (Js). It plays a crucial role in the equation \( E = \frac{hc}{\lambda} \), allowing us to calculate the energy of photons given their wavelength.
Planck's constant encapsulates how quantum systems differ from classical ones. It implies that energy changes occur in discrete amounts, or "quanta." Every photon carries energy that is a multiple of this constant. Thus, Planck's constant is not just a number but a gateway to understanding the quantum nature of the universe.
This concept has led to significant scientific advancements, including the development of technologies like lasers and quantum computing. Therefore, knowing about Planck's constant is essential for anyone diving into the world of quantum physics and photon energy calculations.

Photon Energy Calculation

Calculating photon energy is a crucial skill when dealing with light and its interactions. To find the energy of a photon, we use the formula \( E = \frac{hc}{\lambda} \), where:

• \( E \) is the energy of the photon
• \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{Js}) \)
• \( c \) is the speed of light \( (3 \times 10^8 \text{m/s}) \)
• \( \lambda \) is the wavelength

To get the energy, you substitute the values of Planck's constant and the speed of light, alongside your specific wavelength value into the equation.
In our original exercise, for example, we calculated the energy of photons at a wavelength of \(4000 \) Å from an initial known energy at \(6000 \) Å. By understanding that photons with a shorter wavelength (like \( 4000 \) Å) have higher energy than those at \( 6000 \) Å, and using the inverse proportionality formula, we calculated the requisite energy. This exercise illuminated the practicality of photon energy calculations, essential in fields like spectroscopy, astrophysics, and even medical imaging, where understanding energy differences in light's wavelengths can lead to insightful discoveries and applications.