Problem 25 - Chemistry The Central Science In Si Units

Question

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is 413 \(\mathrm{nm} .\) (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{~J} ?\)

Step-by-Step Solution

Answer

(a) 1.95 × 10^{-19} J, (b) 4.81 × 10^{-19} J, (c) 328 nm.

Step 1: Introduce the formula for photon energy

The energy of a photon is given by the equation \(E = h \cdot f\), where \(E\) is the energy, \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ J} \cdot \text{s}\), and \(f\) is the frequency of the photon.

Step 2: Calculate energy for given frequency

Given the frequency \(f = 2.94 \times 10^{14} \text{ s}^{-1}\), use the formula \(E = h \cdot f\). Substitute in Planck's constant and frequency: \[E = (6.63 \times 10^{-34} \text{ J} \cdot \text{s}) \cdot (2.94 \times 10^{14} \text{ s}^{-1}) = 1.95 \times 10^{-19} \text{ J}\]

Step 3: Introduce the relation between wavelength and energy

The energy of a photon can also be calculated using the wavelength \(\lambda\) with the formula: \[E = \frac{h \cdot c}{\lambda}\]where \(c\) is the speed of light \(3 \times 10^{8} \text{ m/s}\) and \(\lambda\) is the wavelength in meters.

Step 4: Calculate energy from given wavelength

Convert the wavelength from nanometers to meters: \(413 \text{ nm} = 413 \times 10^{-9} \text{ m}\). Use the formula \(E = \frac{h \cdot c}{\lambda}\) to find the energy:\[E = \frac{(6.63 \times 10^{-34} \text{ J} \cdot \text{s}) \cdot (3 \times 10^{8} \text{ m/s})}{413 \times 10^{-9} \text{ m}} = 4.81 \times 10^{-19} \text{ J}\]

Step 5: Find wavelength from given energy

For a photon with energy \(E = 6.06 \times 10^{-19} \text{ J}\), use the formula for wavelength: \[\lambda = \frac{h \cdot c}{E}\]. Substitute \(h\), \(c\), and \(E\):\[\lambda = \frac{(6.63 \times 10^{-34} \text{ J} \cdot \text{s}) \cdot (3 \times 10^{8} \text{ m/s})}{6.06 \times 10^{-19} \text{ J}} = 3.28 \times 10^{-7} \text{ m}\] or equivalently, \(328 \text{ nm}\).

Key Concepts

Planck's Constant

The concept of Planck's constant is central to understanding photon energy calculations. Max Planck, a pioneering physicist, introduced this concept to explain blackbody radiation. Planck's constant, denoted by the symbol \( h \), is a fundamental constant whose value is \( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \). It acts as a bridge between the energy of a photon and its frequency.

This constant is crucial in the formula \( E = h \cdot f \), where \( E \) represents the energy of the photon and \( f \) is the frequency of the electromagnetic wave associated with the photon. Essentially, Planck's constant ensures that energy quantization occurs, meaning energy is only absorbed or emitted in discrete amounts rather than continuous flow.

Understanding Planck's constant allows us to calculate the energy of photons when the frequency is known. It also underscores the foundation of quantum mechanics, reflecting how the universe dictates precise energy exchanges at the microscopic level.

Frequency and Wavelength Relationship

Frequency and wavelength are two crucial aspects of electromagnetic waves, and they are inversely related to each other. This means if one increases, the other decreases. The equation that relates them is given by \( c = \lambda \cdot f \), where \( c \) is the speed of light (\(3 \times 10^{8} \text{ m/s}\)), \( \lambda \) represents the wavelength, and \( f \) is the frequency.

Wavelength is the distance between successive crests of a wave, while frequency is the number of waves that pass a point in one second. These together define the nature and properties of different types of electromagnetic radiation. When you use the formula \( E = \frac{h \cdot c}{\lambda} \) for calculations involving wavelength, you are implicitly using the relationship between frequency and wavelength.

Recognizing this relationship helps in determining not just the type of electromagnetic radiation we are dealing with, but also its behavior and interaction with materials and devices.

Electromagnetic Radiation

Electromagnetic radiation encompasses a broad range of waves that carry energy through space. This includes visible light, radio waves, X-rays, and more. All these waves can be characterized by their wavelength and frequency, which determine their energy levels.

Photons are the elementary particles of electromagnetic waves, acting as quantized packets of energy. According to quantum theory, electromagnetic radiation propagates in waves but interacts with matter in discrete packets or quanta, which is where photon energy calculations come into play.

The energy of a photon is directly related to its frequency and inversely related to its wavelength. Higher-frequency radiation, like X-rays, has higher photon energy compared to lower-frequency radiation, like radio waves. Utilizing the equations \( E = h \cdot f \) and \( E = \frac{h \cdot c}{\lambda} \), we can calculate the energy of photons based on either waveform characteristics.

This understanding is essential for applications ranging from wireless communications, medical imaging, to understanding atomic and molecular structures in physics and chemistry.