Question

A certain molecule has an energy level diagram for its vibrational energy in which two levels are \(0.0141 \mathrm{eV}\) apart. The wavelength of the emitted line for the molecule as it falls from one of these levels to the other, is about (a) \(88 \mu \mathrm{m}\) (b) \(88 \mathrm{~mm}\) (c) \(174.84 \mathrm{~m}\) (d) \(88 \mathrm{~nm}\)

Short answer

The correct answer is (a) \(88 \mu \mathrm{m}\).

Step-by-step solution

- Understand the energy transition relationship to wavelength

To determine the wavelength of the emitted line, we first use the energy difference between the two levels. Recall that the energy of a photon is related to its wavelength by the equation, \( E = \frac{hc}{\lambda} \), where \( E \) is the energy in electron volts (eV), \( h \) is the Planck constant \( (6.626 \times 10^{-34} \mathrm{Js}) \), \( c \) is the speed of light in a vacuum \( (3 \times 10^8 \mathrm{m/s}) \), and \( \lambda \) is the wavelength of the photon in meters.

- Calculate the wavelength associated with the energy transition

With the energy difference \( \Delta E = 0.0141 \mathrm{eV} \) we can find the wavelength \( \lambda \) using the formula \( \lambda = \frac{hc}{\Delta E} \). Remember to convert the energy from electron volts to joules using the conversion \( 1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J} \). So the energy difference in joules is \( \Delta E = 0.0141 \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{J/eV} \).

- Perform the conversion and calculation

Substitute the values into the wavelength equation: \( \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.0141 \times 1.602 \times 10^{-19}} \). Perform the calculation to determine the wavelength in meters, then convert meters to the appropriate unit that matches one of the given options.

- Choose the correct option

Once you've calculated the wavelength in meters, compare your result to the options given in the problem to identify the correct answer.

Key concepts

Energy Transition in Molecules

Molecules exist in various energy states, and transitions between these states can involve the absorption or emission of energy.

When a molecule transitions from a higher to a lower vibrational energy level, it emits energy in the form of a photon. This transition essentially involves quantized energy levels, where the energy difference \( \Delta E \) between two levels is specific to the molecular structure and the type of vibrational change involved.

The emitted photon corresponds to this energy difference, and this phenomenon is crucial in spectroscopy, allowing us to study the properties of molecules. Understanding these energy transactions helps explain the behavior of molecules in various chemical processes and under different environmental conditions. Spectroscopic techniques that utilize this knowledge are fundamental in both analytical and synthetic chemistry.

Photon Energy and Wavelength Relationship

The relationship between photon energy and wavelength is rooted in the fundamental physics of light. According to quantum mechanics, light exhibits both wave-like and particle-like properties.

The energy of a photon \(E\) is inversely proportional to its wavelength \(\lambda\), following the formula \( E = \frac{hc}{\lambda} \) where \(h\) is the Planck constant, and \(c\) is the speed of light.

Therefore, a higher energy photon will have a shorter wavelength, and conversely, a lower energy photon will have a longer wavelength. This concept is vital in fields such as astronomy, where observing the energy of light from stars allows us to determine their composition and other physical properties.

Planck Constant

The Planck constant \(h\) is a fundamental physical constant that plays a pivotal role in quantum mechanics. It relates the energy of a photon to its frequency and has a value of approximately \(6.626 \times 10^{-34} \mathrm{Js}\).

This constant is essential for understanding the quantized nature of energy in the microscopic world. Its existence implies that energy can only be transferred in discrete amounts, rather than any arbitrary value, signifying a departure from classical physics which treated energy as continuous. The Planck constant is not only vital for calculating photon energy but also serves as the foundation for Heisenberg's uncertainty principle and the quantization of the electromagnetic field.

Calculation of Wavelength from Energy Difference

Calculating the wavelength from the energy difference involves rearranging the photon energy and wavelength relationship formula to solve for \(\lambda\). This change gives us \( \lambda = \frac{hc}{\Delta E} \) where \(\Delta E\) is the energy difference between the two states.

To perform the calculation, it is crucial to convert the energy difference from electron volts (eV) to joules (J), using the conversion \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\). After substituting the known values for \(h\), \(c\), and \(\Delta E\), you can calculate the wavelength in meters. Finally, you can convert meters to any required unit, finding the wavelength that corresponds to a given energy transition. This calculation is a cornerstone in understanding spectroscopy and the interaction of light with matter.