Problem 2 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The yellow colour of sodium emission in a flame is due to emission at \(589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm}\). Calculate the difference in energy between the photons emitted at these two wavelengths. (Section 10.2 )

Step-by-Step Solution

Answer

The energy difference is \(3.0 \times 10^{-22} \ \mathrm{J}\).

Step 1: Identify the Given Information

The wavelengths provided are \( \lambda_1 = 589.0 \ \mathrm{nm} \) and \( \lambda_2 = 589.6 \ \mathrm{nm} \). We need to calculate the energy difference between photons emitted at these two wavelengths.

Step 2: Convert Wavelengths to Meters

Convert the given wavelengths from nanometers to meters. Recall that \(1 \mathrm{nm} = 1 \times 10^{-9} \mathrm{m}\). Therefore, \( \lambda_1 = 589.0 \times 10^{-9} \ \mathrm{m} \) and \( \lambda_2 = 589.6 \times 10^{-9} \ \mathrm{m} \).

Step 3: Use the Energy Formula for Photons

The energy of a photon is given by the formula:\[E = \frac{hc}{\lambda}\]where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \ \mathrm{J}\cdot\mathrm{s}) \) and \( c \) is the speed of light \( (3.00 \times 10^8 \ \mathrm{m/s}) \).

Step 4: Calculate Energy for Each Wavelength

Plug in \( \lambda_1 \) and \( \lambda_2 \) into the energy equation:\[E_1 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{589.0 \times 10^{-9}}\]\[E_2 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{589.6 \times 10^{-9}}\]

Step 5: Compute the Energy Values

Calculate:\[ E_1 = 3.371 \times 10^{-19} \ \mathrm{J} \] and \[ E_2 = 3.368 \times 10^{-19} \ \mathrm{J} \] when both calculations are executed.

Step 6: Find the Difference in Energy

Subtract the smaller energy from the larger:\[\Delta E = E_1 - E_2 = 3.371 \times 10^{-19} \ \mathrm{J} - 3.368 \times 10^{-19} \ \mathrm{J} = 3.0 \times 10^{-22} \ \mathrm{J}.\]

Step 7: Final Step: Interpret the Result

The difference in energy between the photons emitted at these wavelengths is small, indicating the close similarity of energy levels that emitted this light.

Key Concepts

Wavelength Conversion

In the realm of physics and chemistry, wavelengths are commonly measured in nanometers (nm). However, when it comes to calculations involving energy, it’s crucial to convert these measurements into meters (m) for consistency with other physical constants. This conversion is simple:

• Remember that 1 nanometer equals \(1 \times 10^{-9}\) meters.
• Thus, a wavelength of 589.0 nm can be converted to meters by multiplying: \(589.0 \times 10^{-9}\).
• Similarly, 589.6 nm becomes \(589.6 \times 10^{-9}\) meters.

By converting nanometers to meters, equations involving Planck’s constant and the speed of light become compatible. This ensures you can accurately calculate the energy of photons.

Planck's Constant

Planck's constant \((h)\) is a fundamental constant in quantum mechanics. It relates the energy of a photon to its frequency. The value of Planck's constant is approximately \(6.626 \times 10^{-34} \) joule seconds (J·s). It is crucial for calculating photon energy:

• The formula \(E = \frac{hc}{\lambda}\) shows that photon energy \(E\) is directly proportional to both Planck's constant \(h\) and the speed of light \(c\).
• As \(h\) is constant, changes in energy primarily depend on variations in wavelength \(\lambda\).

This constant indicates that energy is quantized in discrete units or quanta. Understanding Planck's constant helps in appreciating how light and matter interact at a fundamental level.

Speed of Light

The speed of light \((c)\) is another fundamental constant, which indicates how fast light travels in a vacuum. It is approximately \(3.00 \times 10^8\) meters per second (m/s). This constant plays a pivotal role when calculating the energy of photons:

• Using the formula \(E = \frac{hc}{\lambda}\), the speed of light \(c\) and Planck's constant \(h\) together enable the computation of energy via the wavelength \(\lambda\).
• The constancy of the speed of light ensures that it remains a fixed value, making calculations across different wavelengths consistent.

By combining it with other constants, you can determine the exact amount of energy associated with specific wavelengths of light.

Energy Difference Calculation

Calculating the difference in photon energy is critical in understanding the nuances of different wavelengths. In the exercise, you find this by taking these steps:

• First, calculate the energy for each wavelength using \(E = \frac{hc}{\lambda}\).
• Each calculation gives you an energy value: \(E_1\) for 589.0 nm and \(E_2\) for 589.6 nm.
• Subtract the smaller energy from the larger: \(\Delta E = E_1 - E_2\).

This computation reveals the energy difference, expressed here as \(3.0 \times 10^{-22}\) joules. It highlights how slight changes in wavelength can result in measurable energy differences, helping you understand the quantum nature of light emission.