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A horizontal beam of laser light of wavelength \(585 \mathrm{nm}\) passes through a narrow slit that has width \(0.0620 \mathrm{~mm}\). The intensity of the light is measured on a vertical screen that is \(2.00 \mathrm{~m}\) from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

Short Answer

Expert verified
The minimum uncertainty in the vertical component of the photon's momentum is \(8.50 \times 10^{-28} kg \cdot m/s\) after passing through the slit. The width of the central diffraction maximum on the screen is approximately \(3.6 cm\).

Step by step solution

01

Heisenberg's Uncertainty Principle

Heisenberg's Uncertainty Principle states that we cannot precisely measure both position and momentum of a particle at the same moment. Mathematically it's represented by the inequality \(\Delta p \Delta x \geq \frac{\hbar}{2}\) where \(\Delta p\) denotes the uncertainty in momentum, \(\Delta x\) is the uncertainty in position, and \(\hbar\) is the reduced Planck constant.
02

Calculate Uncertainty in Momentum

Given, the uncertainty in position is the width of the slit, which is \(0.0620 \times 10^{-3} m\). We rearrange the uncertainty principle equation to solve for \(\Delta p\): \( \Delta p \geq \frac{\hbar}{2\Delta x}\) . Plugging the values into the above equation, we find \( \Delta p \geq \frac{(1.055 \times 10^{-34} Js)}{2\times(0.0620 \times 10^{-3} m)} = 8.50 \times 10^{-28} kg \cdot m/s \)
03

Calculate Width of Central Diffraction Maximum

The uncertainty in momentum calculated corresponds to the angle of diffraction in the central maximum (\(\Delta \theta\)) for a single slit, denoted by \( \Delta \theta = \frac{\Delta p}{p} = \frac{\Delta p}{\frac{h}{\lambda}}\), where \(h\) is Planck's constant and \(\lambda\) is the wavelength of the given light. Substituting the given values, we get \( \Delta \theta = \frac{(8.50 \times 10^{-28} kg \cdot m/s)}{(6.626 \times 10^{-34} Js)/(585 \times 10^{-9} m)} = 8.95 \times 10^{-3} rad\). The width of the central maximum on the screen is given by \(w= 2L \Delta \theta\), where \(L\) is the distance from the slit to the screen. Substituting the given values, we get: \(w = 2 \times (2 m) \times (8.95 \times 10^{-3} rad) = 0.036 m = 3.6 cm\). That is the width of the central maximum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heisenberg's Uncertainty Principle
Understanding quantum mechanics often begins with Heisenberg's Uncertainty Principle. This fundamental concept suggests that we can't precisely determine both the position and momentum of a particle simultaneously. This principle is mathematically expressed as:\[ \Delta p \Delta x \geq \frac{\hbar}{2} \]Here, \(\Delta p\) stands for the uncertainty in momentum and \(\Delta x\) for the uncertainty in position. The symbol \(\hbar\) represents the reduced Planck constant, which is a fundamental constant in quantum mechanics.In our exercise, the slit width translates directly to the position uncertainty \(\Delta x\). It quantifies how precisely we can say where the photon is as it passes through the slit. By knowing this, we can then determine the corresponding minimum uncertainty in the photon's momentum, giving us an insight into its behavior post-filtration through the slit.
Photon Momentum
Photon momentum might not be something we think about often, but light particles carry momentum even without mass. In physics, the momentum \(p\) of a photon is given by:\[ p = \frac{h}{\lambda} \]Where \(h\) is Planck's constant and \(\lambda\) is the photon's wavelength. This equation shows that the momentum of a photon is inversely proportional to its wavelength—shorter wavelengths mean more momentum.In this case, after applying Heisenberg's principle, we find the uncertainty in this photon momentum. \[ \Delta p \geq \frac{\hbar}{2\Delta x} \]That tells us how much the momentum might waver as it squeezes through the narrow slit, causing it to spread out in different directions.
Central Diffraction Maximum
When light interacts with obstacles, it bends slightly, creating patterns. This is known as diffraction. The central diffraction maximum refers to the brightest spot on a screen where light bands form, straight opposite the opening causing the diffraction.In our task, after calculating the uncertainty in photon momentum, this \(\Delta p\) is linked to the angular spread \(\Delta \theta\) of light in the central diffraction maximum:\[ \Delta \theta = \frac{\Delta p}{p} \]Returning to simple geometry, the width \(w\) of this central band on the screen is calculated as:\[ w = 2L \Delta \theta \]Where \(L\) is the distance from slit to screen. This width accounts for the range over which the majority of the light intensity is concentrated and is an essential detail for understanding how different variables affect the diffraction pattern.

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Most popular questions from this chapter

A photon with wavelength \(\lambda=0.1385 \mathrm{nm}\) scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is \(8.90 \times 10^{6} \mathrm{~m} / \mathrm{s} ?\)

When ultraviolet light with a wavelength of \(400.0 \mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be \(1.10 \mathrm{eV}\). What is the maximum kinetic energy of the photoelectrons when light of wavelength \(300.0 \mathrm{nm}\) falls on the same surface?

A photon with wavelength \(0.1100 \mathrm{nm}\) collides with a free electron that is initially at rest. After the collision the wavelength is \(0.1132 \mathrm{nm}\). (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

A 75 W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength \(600 \mathrm{nm}\). (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

We can estimate the number of photons in a room using the following reasoning: The intensity of the sun's rays at the earth's surface is roughly \(1000 \mathrm{~W} / \mathrm{m}^{2}\). A typical room is illuminated by indirect light or by light bulbs so that its light intensity is some fraction of that value. (a) Estimate the intensity of the light that enters your room. (b) Model the light as entering uniformly at the ceiling and exiting uniformly at the floor. Estimate the area \(A\) of the floor and ceiling, and the height \(H\) of your room. (c) Estimate how long it takes light to travel from the ceiling to the floor of your room by dividing \(H\) by the speed of light. (d) Estimate the total power of the light that enters your room \(P\) by multiplying your estimated intensity \(I\) by the area of your ceiling \(A\). (e) Estimate the total light energy in your room \(E_{\text {room }}\) by multiplying the power \(P\) by the length of time it takes light to travel from ceiling to floor. (f) An average wavelength for light is in the middle of the visible spectrum, at roughly \(500 \mathrm{nm}\). What is the energy of a \(500 \mathrm{nm}\) photon? (g) The total number of photons in your room \(N\) is the ratio of the energy \(E_{\text {room }}\) to the energy per photon. What is your estimate for \(N ?\)

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