Question

Rank these photons in terms of increasing energy: blue \((\lambda=\) \(453 \mathrm{nm}) ;\) red \((\lambda=660 \mathrm{nm}) ;\) yellow \((\lambda=595 \mathrm{nm})\)

Short answer

Red < Yellow < Blue

Step-by-step solution

Understand the relationship between wavelength and energy

Photon energy is inversely proportional to wavelength. The formula to calculate the energy of a photon is given by \[ E = \frac{hc}{\text{λ}} \] where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{J·s}\), \(c\) is the speed of light \(3 \times 10^{8} \text{m/s}\), and \(λ\) is the wavelength of light.

Calculate the energy for each photon

Using the equation \[ E = \frac{hc}{\text{λ}} \], calculate the energy for each wavelength: For blue photons (λ=453 nm): \[ E_{\text{blue}} = \frac{(6.626 \times 10^{-34} \text{J·s})(3 \times 10^{8} \text{m/s})}{453 \times 10^{-9} \text{m}} \]For red photons (λ=660 nm): \[ E_{\text{red}} = \frac{(6.626 \times 10^{-34} \text{J·s})(3 \times 10^{8} \text{m/s})}{660 \times 10^{-9} \text{m}} \]For yellow photons (λ=595 nm): \[ E_{\text{yellow}} = \frac{(6.626 \times 10^{-34} \text{J·s})(3 \times 10^{8} \text{m/s})}{595 \times 10^{-9} \text{m}} \]

Compare the energies

Compare the calculated energies to rank the photons. Since energy is inversely proportional to wavelength, the photon with the longest wavelength (red) will have the least energy and the one with the shortest wavelength (blue) will have the most energy.

Rank the photons in terms of increasing energy

Based on the wavelengths given (red=660 nm, yellow=595 nm, blue=453 nm), rank them as follows:1. Red photon 2. Yellow photon 3. Blue photon.

Key concepts

wavelength and energy relationship

Photons exhibit an intriguing property: their energy is inversely proportional to their wavelength. This means that as the wavelength \( \text{λ} \) of a photon becomes shorter, its energy \( \text{E} \) increases, and vice versa. The exact relationship between a photon's wavelength and its energy can be described by the formula: \[ E = \frac{hc}{\text{λ}} \].
We can break this down into simpler terms:

• \( \text{E} \) represents the energy of the photon.
• \( h \) denotes Planck's constant, equivalent to \( 6.626 \times 10^{-34} \text{J·s} \).
• \( c \) is the speed of light, which is \( 3 \times 10^{8} \text{m/s} \).
• \( \text{λ} \) stands for the wavelength of the photon.

Using this formula, we see that a shorter wavelength results in a larger value for \( \frac{1}{\text{λ}} \), which therefore boosts the energy \( \text{E} \). Conversely, a longer wavelength reduces \( \frac{1}{\text{λ}} \) and thus the photon's energy.
Consider the colors of light: blue light has a shorter wavelength than red light, which is why blue photons are more energetic than red photons.

Planck's constant

Planck's constant, symbolized as \( h \), plays a crucial role in quantum mechanics and the calculation of photon energy. It has a value of \( 6.626 \times 10^{-34} \text{J·s} \). Planck's constant essentially quantifies the smallest possible unit of energy transition in quantum systems. In the context of photon energy, it helps correlate the energy of a photon to its frequency and wavelength through the formula \[ E = \frac{hc}{\text{λ}} \].
This constant underscores the discrete nature of energy levels:

• Energy is quantized, meaning it can only exist in specific discrete amounts rather than a continuous range.
• Every photon—a particle of light—carries energy that is directly tied to its frequency, dictated by Planck's constant.

By discovering Planck's constant, we gained a powerful tool to understand and calculate various phenomena in quantum physics, especially those involving light and photons.

speed of light

The speed of light, denoted as \( c \), is a fundamental constant in physics—it is the speed at which light travels in a vacuum and is valued at \( 3 \times 10^{8} \text{m/s} \). This constant is key to understanding wave-based phenomena and is especially important when calculating the energy of photons using the formula: \[ E = \frac{hc}{\text{λ}} \].
Here's why \( c \) matters:

• It is a constant value in the universe, meaning it does not change regardless of the observer's frame of reference.
• It directly relates the frequency and wavelength of any electromagnetic wave, including light photons.
• The high value of \( 3 \times 10^{8} \text{m/s} \) demonstrates just how fast light travels, aiding in precise calculations of energy and other wave properties.

With \( c \), we can fully comprehend how photons, which are fundamental light particles, carry energy and exhibit properties of both waves and particles—a duality central to quantum mechanics.