Problem 7 - Chemistry and Chemical Reactivity

Question

The most prominent line in the emission spectrum of aluminum is at \(396.15 \mathrm{nm}\). What is the frequency of this line? What is the energy of one photon with this wavelength? Of 1.00 mol of these photons?

Step-by-Step Solution

Answer

The frequency is approximately \(7.57 \times 10^{14} \text{ Hz}\). The energy of one photon is approximately \(5.01 \times 10^{-19} \text{ J}\), and 1 mol of photons is approximately \(3.02 \times 10^{5} \text{ J/mol}\).

Step 1: Convert Wavelength to Frequency

To find the frequency of the line, we use the formula that relates frequency \( f \), wavelength \( \lambda \), and the speed of light \( c \): \[ c = \lambda \times f \]Here, \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and the wavelength \( \lambda = 396.15 \text{ nm} = 396.15 \times 10^{-9} \text{ m} \).Rearrange the formula to solve for \( f \):\[ f = \frac{c}{\lambda} \]Substitute the known values:\[ f = \frac{3.00 \times 10^8}{396.15 \times 10^{-9}} \approx 7.57 \times 10^{14} \text{ Hz} \]

Step 2: Calculate Energy of One Photon

To find the energy of one photon, we use Planck's equation:\[ E = h \times f \]where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J·s} \), and \( f \approx 7.57 \times 10^{14} \text{ Hz} \) from Step 1.Substitute the values:\[ E = 6.626 \times 10^{-34} \times 7.57 \times 10^{14} \approx 5.01 \times 10^{-19} \text{ J} \]

Step 3: Calculate Energy of 1.00 mol of Photons

Using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \), we find the total energy for 1 mol of these photons.The energy of 1 molecule is \( 5.01 \times 10^{-19} \text{ J} \) from Step 2, so for 1 mol:\[ E_{\text{1 mol}} = 5.01 \times 10^{-19} \times 6.022 \times 10^{23} \approx 3.02 \times 10^{5} \text{ J/mol} \]

Key Concepts

Frequency

Understanding frequency is key when delving into the concepts of light and emission spectra. Frequency (nu) is defined as the number of waves that pass a fixed point in one second. In simpler terms, it reflects how often the wave "wiggles" in a given timeframe.

The relationship between wavelength (lambda) and frequency is inversely proportional. That means as one increases, the other decreases. For light, this relationship is bound by the speed of light (c), roughly equal to 3.00 \(\times\) 10\(^8\) m/s. Thus, the formula connecting these quantities is:

• \[ c = \lambda \times u \]

To get the frequency when you know the wavelength, you'd rearrange the formula to:

• \[ u = \frac{c}{\lambda} \]

By substituting given values, like the problem’s 396.15 nm (or 396.15 \(\times\) 10\(^{-9}\) m in meters), you can calculate the frequency easily. This is essential in understanding how different wavelengths in the emission spectrum correspond to different energies and characteristics of light.

Photon Energy

Photon energy relates closely to frequency and is defined through Planck's equation, which connects these aspects elegantly. This equation allows us to determine the energy carried by a single photon if its frequency is known.

The formula is:

• \[ E = h \times u \]

where \( E \) is energy, \( h \) is Planck’s constant (approximately 6.626 \(\times\) 10\(^{-34}\) J·s), and \( u \) is frequency. Given frequency from the prior steps, calculating photon energy becomes straightforward.

In practical applications, this means that photons with higher frequencies carry more energy. This is why ultraviolet light, known for its high frequency, carries more energy than visible light. Each color of light in the spectrum aligns with a specific frequency and a corresponding energy level, defining its characteristics and effects.

Mole Calculations

In chemistry, the mole is a fundamental unit that measures amounts of substance. Avogadro's number, \( N_A \), plays a vital role in this calculation. It provides the conversion factor from the number of entities (like photons or molecules) to moles, where \( N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1} \).

To find the total energy associated with a mole of photons, you multiply the energy of one photon by Avogadro’s number:

• \[ E_{\text{1 mol}} = E \times N_A \]

This calculation is vital in fields like quantum chemistry and materials science, where the energy of large quantities of particles needs to be considered. So, if one photon of a particular energy contributes to a reaction, calculating the effect of a mole of these photons helps gauge the energy changes and potential transformations that can occur.