Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

If the wavelength of an electromagnetic radiation is \(2000 \AA\). What is the energy in ergs? (a) \(9.92 \times 10^{-19}\) (b) \(9.94 \times 10^{-12}\) (c) \(4.97 \times 10^{-12}\) (d) \(4.97 \times 10^{-19}\)

Step-by-Step Solution

Answer

The energy is approximately \( 9.94 \times 10^{-12} \) ergs, option (b).

Step 1: Understand the Problem

We are given the wavelength of electromagnetic radiation as 2000 Ångströms and need to find the energy of this radiation in ergs. We will use the equation that relates energy (E) to wavelength (λ): \[ E = \frac{hc}{\lambda} \]where \( h \) is Planck's constant \( (6.63 \times 10^{-27} \text{ ergs}) \) and \( c \) is the speed of light \( (3 \times 10^{10} \text{ cm/s}) \). Note that 1 Ångström is \( 1 \times 10^{-8} \text{ cm} \).

Step 2: Convert Wavelength to Centimeters

The wavelength is given as 2000 Ångströms. First, convert this to centimeters:\[ 2000 \text{ Å} = 2000 \times 10^{-8} \text{ cm} = 2 \times 10^{-5} \text{ cm} \]

Step 3: Calculate Energy Using the Formula

Now substitute the known values into the energy formula:\[ E = \frac{(6.63 \times 10^{-27} \text{ erg\cdot s})(3 \times 10^{10} \text{ cm/s})}{2 \times 10^{-5} \text{ cm}} \]Calculate the energy:\[ E = \frac{1.989 \times 10^{-16} \text{ erg\cdot cm/s}}{2 \times 10^{-5} \text{ cm}} = 9.945 \times 10^{-12} \text{ erg} \]

Step 4: Round the Energy to Match Options

Review the calculated energy and compare it to the given options. Rounding the energy result, we see that it is approximately \( 9.94 \times 10^{-12} \) ergs, which corresponds to option (b).

Key Concepts

Electromagnetic Spectrum

The electromagnetic spectrum includes all types of electromagnetic radiation, ranging from very low-energy radio waves to extremely high-energy gamma rays. Each type of radiation is categorized by its wavelength or frequency. This spectrum is crucial for understanding how different kinds of electromagnetic waves work and how they affect matter.

When you think of visible light, you're considering just a small portion of the electromagnetic spectrum. Invisible spectrums like ultraviolet (UV) and infrared (IR) waves often go unnoticed but play significant roles in various technologies and natural processes.

• Radio waves are used in communication devices like radios and televisions.
• Micro-waves are used in cooking (microwaves) and in radar technology.
• Infrared is used in heaters and night-vision technologies.
• X-rays are commonly used in medical imaging.
• Gamma rays are often used in cancer treatment.

The further you go from visible light on the spectrum, the more specialized the uses of the radiation become. Understanding where a specific wavelength falls on this spectrum helps us determine its potential applications and effects.

Wavelength Conversion

Wavelength conversion is a necessary step in many scientific calculations, particularly when working with electromagnetic waves. In the original exercise, the wavelength was given in Ångströms, a unit often used in atomic physics to measure atomic and molecular scales.

To use the energy calculation formula properly, you need to convert this wavelength into centimeters, because the constants involved are typically measured in centimeters.

• 1 Ångström is equal to \(1 \times 10^{-8}\) centimeters
• This conversion ensures consistency and avoids computation errors

Let's take the current example: 2000 Ångströms.
By converting this into centimeters:
\[ 2000 \text{ Å} = 2000 \times 10^{-8} \text{ cm} = 2 \times 10^{-5} \text{ cm} \]
This is a straightforward conversion, but it's critical for ensuring accurate work with the formulas that don't naturally accommodate the Ångström unit.

Planck's Constant

Planck's constant is a fundamental constant in physics that describes the quantization of energy. It essentially tells us how the energy of radiation is quantized, implying that energy comes in discrete packets, or quanta. Planck’s constant \( (h) \) is approximately \( 6.63 \times 10^{-27} \text{ erg\cdot s} \).

The introduction of Planck's constant was revolutionary in understanding the nature of electromagnetic radiation and led to the development of quantum mechanics. In the given energy calculation formula, Planck's constant is vital because:

• It multiplies with the speed of light and then divides by the wavelength to give the photon energy
• In many equations, having the constant allows us to relate the frequency and energy of photons
• It provides a bridge between wave (wavelength) and particle (energy) descriptions of radiation

Without Planck's constant, our current understanding of many physical phenomena would be incomplete.

Speed of Light

The speed of light, denoted as \(c\), is another fundamental constant in physics. This value is approximately \(3 \times 10^{10} \text{ cm/s}\). This speed represents how fast light travels in a vacuum and is a constant in all equations involving electromagnetic radiation.

In calculations involving energy and electromagnetic radiation:

• The speed of light helps determine how quickly light waves propagate through space
• It is essential for equations that determine energy from wavelength, as it works alongside Planck's constant
• In the discussed formula, multiplying the speed of light with Planck's constant helps calculate the energy of a photon based on wavelength

Given its fundamental nature, the speed of light is not just critical in physics but also in technologies like GPS and wireless communication. This constant underlines the interconnectedness of various physical laws and constants.