Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

The energy ratio of a photon of wavelength \(3000 \AA\) and \(6000 \AA\) is (a) \(1: 1\) (b) \(2: 1\) (c) \(1: 2\) (d) \(1: 4\)

Step-by-Step Solution

Answer

The energy ratio of a photon of wavelength 3000 Å and 6000 Å is 2:1.

Step 1: Understanding Photon Energy Formula

The energy of a photon is calculated using the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant \( (6.63 \times 10^{-34} \, Js)\), \(c\) is the speed of light \((3 \times 10^8 \, m/s)\), and \(\lambda\) is the wavelength.

Step 2: Calculate Energy for Wavelength 3000 Å

Convert the wavelength from angstroms to meters: \(3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m}\). Substitute \(\lambda = 3 \times 10^{-7} \, \text{m}\) into the energy formula: \(E_{3000} = \frac{6.63 \times 10^{-34} \, Js \times 3 \times 10^8 \, m/s}{3 \times 10^{-7} \, m}\). Simplifying gives \(E_{3000} = \frac{6.63 \times 3}{3} \times 10^{-19} \, J = 6.63 \times 10^{-19} \, J\).

Step 3: Calculate Energy for Wavelength 6000 Å

Convert the wavelength from angstroms to meters: \(6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m}\). Substitute \(\lambda = 6 \times 10^{-7} \, \text{m}\) into the energy formula: \(E_{6000} = \frac{6.63 \times 10^{-34} \, Js \times 3 \times 10^8 \, m/s}{6 \times 10^{-7} \, m}\). Simplifying gives \(E_{6000} = \frac{6.63 \times 3}{6} \times 10^{-19} \, J = 3.315 \times 10^{-19} \, J\).

Step 4: Calculate the Energy Ratio

Now, compare the energies calculated: \(\frac{E_{3000}}{E_{6000}} = \frac{6.63 \times 10^{-19}}{3.315 \times 10^{-19}} = 2\). Thus, the energy ratio is \(2:1\).

Key Concepts

Wavelength Conversion

When dealing with photons and calculating their energy, wavelengths are commonly given in various units such as angstroms (Å). In the scientific world, it's crucial to convert these units into meters for consistency and compatibility with other constants, like the speed of light and Planck's constant, which are typically expressed in terms of meters. To convert wavelengths from angstroms to meters, remember that:

• 1 angstrom (Å) is equal to \(10^{-10}\) meters.

So, for example, a photon with a wavelength of 3000 Å becomes:

• \(3000 \times 10^{-10}\) meters.

Likewise, a photon with a wavelength of 6000 Å translates to:

• \(6000 \times 10^{-10}\) meters.

These converted values make it easier to use them in formulas that require measurements in meters, helping us calculate photon energy accurately.

Energy Ratio Calculation

Calculating the energy ratio between photons of different wavelengths can be straightforward when following the photon energy formula, which employs Planck’s constant and the speed of light. The important thing to remember is that the energy of a photon is inversely proportional to its wavelength. Therefore, photons with shorter wavelengths have higher energies.To find the energy ratio between two photons, calculate the energy for each, then take the ratio of these two energies. Using the photon energy formula:

• \( E = \frac{hc}{\lambda} \)

we find:

• \( E_{3000} = \frac{6.63 \times 10^{-34} \text{Js} \times 3 \times 10^8 \text{m/s}}{3 \times 10^{-7} \text{m}} \)
• \( E_{6000} = \frac{6.63 \times 10^{-34} \text{Js} \times 3 \times 10^8 \text{m/s}}{6 \times 10^{-7} \text{m}} \)

The calculations yield:

• \( E_{3000} = 6.63 \times 10^{-19} \) Joules
• \( E_{6000} = 3.315 \times 10^{-19} \) Joules

Dividing the energy of the photon with a shorter wavelength by the energy of the photon with a longer wavelength gives us the energy ratio:

• \( \frac{E_{3000}}{E_{6000}} = 2:1 \)

This result highlights the principle that as wavelength decreases, energy increases.

Planck's Constant

Planck’s constant is a fundamental constant that plays a critical role in quantum mechanics, specifically in the calculation of photon energy. This constant represents the proportionality between the energy of a photon and the frequency of its corresponding electromagnetic wave.Denoted as \( h \), Planck's constant has the value:

• \( 6.63 \times 10^{-34} \text{Js} \)

The importance of Planck's constant arises from its application in the expression:

• \( E = hf \) or \( E = \frac{hc}{\lambda} \)

where \( c \) is the speed of light and \( \lambda \) is the photon wavelength. This equation shows that energy is directly proportional to frequency (\( f \)) or inversely proportional to wavelength.Planck’s constant helps us bridge the gap between observed macroscopic phenomena and the quantum mechanics underlying it, providing a foundational component in the study of quantum energy levels and photon behavior. Understanding this concept allows for deeper insight into how energy and wavelength are fundamentally connected.