Question

The dissociation energy of \(\mathrm{N}_{2}\) is very high, \(941 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the wavelength of the photons that possess sufficient energy to dissociate \(\mathrm{N}_{2} .(\mathbf{b})\) In which region of the electromagnetic spectrum does this light fall? Does this light have enough energy to photoionize \(\mathrm{N}_{2}\) ?

Short answer

The wavelength of photons with sufficient energy to dissociate N₂ molecules is 1.271 x 10^(-7) m or 127.1 nm, which falls under the Ultraviolet (UV) region of the electromagnetic spectrum. However, these photons do not have enough energy to photoionize N₂, as the ionization energy of N₂ is approximately 1500 kJ/mol, which is higher than its dissociation energy of 941 kJ/mol.

Step-by-step solution

Convert the dissociation energy to Joules

The dissociation energy is given in kJ/mol, which needs to be converted to Joules (J) to be used in the energy equation. Since 1 kJ = 1000 J, Dissociation energy = 941 kJ/mol × (1000 J / 1 kJ) = 941,000 J/mol

Calculate the energy per photon

To find the energy per photon, we need to divide the dissociation energy by Avogadro's number (6.022 x 10^23 mol^(-1)): Energy per photon = 941,000 J/mol ÷ (6.022 x 10^23 mol^(-1)) = 1.562 x 10^(-19) J

Calculate the wavelength of the photons

Using the energy equation, we can now find the wavelength of the photons: \( \lambda = \dfrac{h \times c}{E} \) \( \lambda = \dfrac{ (6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^{8} \text{ m/s})}{1.562 \times 10^{-19} \text{ J}} \) \( \lambda = 1.271 \times 10^{-7} \text{ m} \)

Determine the region of the electromagnetic spectrum

The wavelength of the photons is 1.271 x 10^(-7) m or 127.1 nm, which falls under the Ultraviolet (UV) region of the electromagnetic spectrum.

Determine if the light has enough energy to photoionize N₂

The ionization energy of N₂ is approximately 1500 kJ/mol, which is higher than the dissociation energy of 941 kJ/mol. Therefore, the photons with a wavelength of 127.1 nm have enough energy to dissociate N₂ but do not have enough energy to photoionize N₂.

Key concepts

Dissociation Energy

When we talk about dissociation energy, we are discussing the energy required to break chemical bonds. For a molecule like \(_2\), which is made up of two nitrogen atoms, dissociation energy is crucial for understanding how much energy is needed to separate these atoms into individual, neutral atoms.

The unit for dissociation energy is often given in kilojoules per mole (kJ/mol). This describes the energy needed to dissociate one mole of a particular substance. In the case of nitrogen, \(_2\), its dissociation energy is 941 kJ/mol. This is a relatively high value compared to other molecules, meaning it takes a lot of energy to break the nitrogen- nitrogen bond.

Understanding dissociation energy is vital an example in atmospheric chemistry and combustion processes, where molecular bonds are broken and formed to release energy or perform chemical reactions.

Electromagnetic Spectrum

The electromagnetic spectrum is a chart that organizes all the different kinds of electromagnetic radiation. These radiations vary in their wavelengths and frequencies, ranging from very small gamma rays to huge radio waves.

One of the key ideas is that the energy of electromagnetic radiation increases as its wavelength decreases. For instance, gamma rays have high energy and short wavelengths, while radio waves have low energy and long wavelengths.

• From shortest to longest wavelength, the spectrum includes gamma rays, X-rays, ultraviolet (UV) radiation, visible light, infrared radiation, microwaves, and radio waves.
• This spectrum is crucial for understanding the behaviors and applications of different types of radiation—from medical imaging with X-rays to communication networks using radio waves.
• Every different part of the spectrum has unique uses and significance in both scientific and everyday contexts.

Wavelength Calculation

Calculating the wavelength of a photon involves using the energy-wavelength relationship provided by Planck's equation. This is a simple yet powerful formula that links energy (E) and wavelength (\( \lambda \)):

\[ \lambda = \dfrac{hc}{E} \]

Where \(h\) is Planck’s constant (6.626 x 10^{-34} J s), \(c\) is the speed of light (3.0 x 10^8 m/s), and \(E\) is the energy of the photon.

In our example, the calculated energy per photon was 1.562 x 10^(-19) J. Substituting these values into the equation, we find:

• \( \lambda = \dfrac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^{8} \text{ m/s})}{1.562 \times 10^{-19} \text{ J}} \)
• This results in a wavelength of 1.271 x 10^(-7) m, or 127.1 nanometers (nm).

This step shows how we can move from knowing the energy of a photon to calculating its wavelength, a key skill when examining the interactions of light with matter.

Ultraviolet Radiation

Ultraviolet (UV) radiation is a type of electromagnetic radiation that falls in between visible light and X-rays on the electromagnetic spectrum.

It is known for having wavelengths between about 10 nm to 400 nm and is characterized by higher energy and shorter wavelengths than visible light.

This high energy makes UV radiation particularly interesting because:

• It is energetic enough to cause chemical reactions, such as dissociation, in certain molecules.
• It's also responsible for effects like skin tanning or burning due to the UV radiation from the sun.
• In our problem example, the UV radiation we calculated is represented by a wavelength of 127.1 nm. This falls conveniently in the UV region, confirming that the photons possess sufficient energy to dissociate \(\mathrm{N}_{2}\) molecules.

The ability of UV radiation to provoke chemical changes is why certain UV lights are used for sterilization in both medical and commercial settings.