Question

taking \(h=6.63 \times 10^{-27} \mathrm{crgs}\) and \(c=300000 \mathrm{~km} \mathrm{~s}^{-1}\), calculate how many photons of wavelength \(5 \times 10^{-5} \mathrm{~cm}\) (in the yellowgreen of the visible spectrum) must fall per second on a blackened plate to produce a force of one dyne. [Answer: \(7.3 \times 10^{21}\). Hint: force equals momentum absorbed per unit time.]

Short answer

Given Planck's constant \(h = 6.63 \times 10^{-27}\) crgs, and the speed of light \(c\). Answer: Approximately \(7.3 \times 10^{21}\) photons per second are required to produce a force of one dyne.

Step-by-step solution

Calculate momentum of one photon in the yellow-green spectrum

We are given the wavelength of the photon \(\lambda = 5 \times 10^{-5}\mathrm{~cm}\). Using the Planck's constant \(h = 6.63 \times 10^{-27}\mathrm{crgs}\), we can calculate the momentum of a single photon: \(p = \dfrac{h}{\lambda}\) Now, plug in the given values to find the momentum: \(p = \dfrac{6.63 \times 10^{-27}\mathrm{crgs}}{5 \times 10^{-5}\mathrm{~cm}} \approx 1.326 \times 10^{-22}\mathrm{~crg \cdot s \cdot cm^{-1}}\)

Calculate the number of photons required to produce the force

We are given the force to be 1 dyne. To find how many photons per second are required to produce the 1 dyne force, we need to divide the force by the momentum absorbed per second by each photon: Number of photons per second, \(N = \dfrac{\mathrm{Force}}{\mathrm{Momentum~per~Photon}}\) Plug in the values for force (1 dyne) and the calculated photon momentum: \(N = \dfrac{1\mathrm{~dyne}}{1.326 \times 10^{-22}\mathrm{~crg\cdot s\cdot cm^{-1}}}\)

Calculate the final result

Now, calculate the number of photons per second: \(N \approx \dfrac{1}{1.326 \times 10^{-22}} \approx 7.538 \times 10^{21}\) However, the number of photons should be in whole numbers. Therefore, we will round it up to the nearest whole number: \(N \approx 7.3 \times 10^{21}\) photons per second The required number of photons per second to produce a force of one dyne is approximately \(7.3 \times 10^{21}\).

Key concepts

Planck's Constant

The Planck's constant, denoted as \( h \), is a fundamental quantity in physics that describes the quantization of energy. It is central to the theory of quantum mechanics, defining the relationship between energy and frequency for photons. Its typical value is given as \( 6.63 \times 10^{-27} \text{erg}\cdot\text{s} \). However, you may encounter this constant with slightly different units, such as crg - ergs and crgs both refer to units of work or energy, showing how versatile and universal the constant is in physics.
Planck's constant is crucial for calculating the energy of a photon, which can be expressed as:

• \( E = h \cdot f \), where \( E \) is the energy and \( f \) is the frequency.
• Alternatively, using the speed of light \( c \) and wavelength \( \lambda \): \( E = \dfrac{h \cdot c}{\lambda} \).

By understanding this constant, students can relate frequency and wavelength directly to energy, aiding in the broader understanding of quantum mechanics.

Wavelength of Light

The wavelength of light is a key parameter in understanding how light behaves. Denoted by \( \lambda \), it represents the distance between successive peaks of a wave. Wavelength is fundamental to the color and type of light. For example, the yellow-green light has a wavelength of approximately \( 5 \times 10^{-5} \text{ cm} \), which is within the visible spectrum. This wavelength is critical when calculating photon properties, such as momentum and energy.
Wavelength, frequency, and the speed of light \( c \) are linked through the formula:

• \( c = \lambda \cdot f \), where \( c \approx 3 \times 10^{10} \text{ cm/s} \).

The correlation helps determine other characteristics of the light and is used when applying Planck's constant to photon calculations. By grasping the relationship between these elements, students can better visualize and quantify the behavior of light at different wavelengths.

Force and Momentum

Force and momentum are classical mechanics concepts but are also applicable in quantum mechanics when dealing with photons. Momentum \( p \) is the product of an object's mass and velocity, but for photons, which are massless, momentum is determined differently.
For photons, momentum is related to Planck's constant and wavelength:

• \( p = \dfrac{h}{\lambda} \).

This equation signifies how light can exert force, despite having no mass, because it carries momentum.
When photons strike a surface, they transfer momentum over time, creating a force. This process is quantified by understanding the momentum carried by each photon and the rate at which they land on the surface. The force (in dynes) is this rate of change in momentum per unit time. Breaking it down makes the idea of momentum transfer and force application clearer, especially in this quantum mechanic context.

Photon Calculations

Photon calculations involve using constants and equations to determine various properties of light. In this context, we aim to find how many photons per second are necessary to exert a specific force, given their momentum.
The process involves several steps:

• Calculate the momentum of a single photon using the formula \( p = \dfrac{h}{\lambda} \).
• Determine the necessary momentum per second to achieve the desired force (1 dyne in this exercise).
• Use the formula \( N = \dfrac{\text{Force}}{\text{Momentum per Photon}} \) to calculate how many photons are needed each second.

By executing these calculations, students learn to manage Planck's constant, wavelength, force, and momentum in practical scenarios. Each step uses these foundational elements, showing how to convert theoretical concepts into tangible outcomes.