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An electron moving with a speed of \(2.7 \times 10^{6} \mathrm{m} / \mathrm{s}\) has the same momentum as a photon. Find (a) the de Broglie wavelength of the electron and (b) the wavelength of the photon.

Short Answer

Expert verified
(a) Wavelength of the electron: \(2.69 \times 10^{-10}\) m; (b) Wavelength of the photon: \(2.69 \times 10^{-10}\) m.

Step by step solution

01

Calculate the Momentum of the Electron

Momentum of the electron (p) is given by the equation: \( p = m \times v \), where \( m \) is the mass of the electron \( (9.11 \times 10^{-31} \text{ kg}) \) and \( v \) is the speed of the electron \( (2.7 \times 10^{6} \text{ m/s}) \). Compute: \[ p = 9.11 \times 10^{-31} \times 2.7 \times 10^{6} = 2.46 \times 10^{-24} \text{ kg m/s} \].
02

Find the de Broglie Wavelength of the Electron

The de Broglie wavelength \( \lambda_{e} \) is given by \( \lambda_{e} = \frac{h}{p} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ m}^2 \text{kg/s}) \) and \( p \) is the momentum calculated in Step 1. Thus, \( \lambda_{e} = \frac{6.626 \times 10^{-34}}{2.46 \times 10^{-24}} = 2.69 \times 10^{-10} \text{ m} \).
03

Equate the Photon's Momentum with Electron's Momentum

Since the photon has the same momentum as the electron, \( p_{\text{photon}} = 2.46 \times 10^{-24} \text{ kg m/s} \).
04

Calculate the Wavelength of the Photon

For the photon, the momentum is given by \( p = \frac{h}{\lambda} \). Equating with electron's momentum: \( \frac{h}{\lambda_{\text{photon}}} = 2.46 \times 10^{-24} \). Solve for \( \lambda_{\text{photon}} \): \( \lambda_{\text{photon}} = \frac{6.626 \times 10^{-34}}{2.46 \times 10^{-24}} = 2.69 \times 10^{-10} \text{ m} \).

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Key Concepts

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

Electron Momentum
When discussing the momentum of an electron, we rely on a classical physics concept. - Momentum is the measure of the motion of an object and is represented by the product of an object's mass and its velocity.- The formula for momentum is: \[ p = m \times v \] where: - \( p \) = momentum, - \( m \) = mass, and - \( v \) = velocity.In our exercise:- The electron has a mass of \( 9.11 \times 10^{-31} \text{ kg} \) and moves with a velocity of \( 2.7 \times 10^{6} \text{ m/s} \).- Substituting these values into the momentum formula, we find the electron's momentum is \( 2.46 \times 10^{-24} \text{ kg m/s} \).This momentum is important because it creates a link between particles with mass like electrons and massless particles like photons. Understanding an electron's momentum is the starting point for exploring deeper quantum mechanics concepts.
Photon Momentum
Photons, unlike electrons, are massless particles, but they also possess momentum.- The momentum of a photon does not rely on mass. Instead, it's connected to the energy and wavelength of the photon.- The formula for a photon's momentum is: \[ p = \frac{h}{\lambda} \] where: - \( p \) = momentum, - \( h \) = Planck's constant, and - \( \lambda \) = wavelength.In this exercise:- We equated the momenta of the electron and the photon since they have been deemed equal.- Thus, the momentum of the photon was also \( 2.46 \times 10^{-24} \text{ kg m/s} \).Using the relationship between momentum and wavelength, we can derive crucial insights into the behavior of particles like photons, even though they lack mass. This demonstrates the wave-particle duality seen in quantum mechanics.
Planck's Constant
Planck's constant is a fundamental constant in physics, essential in the field of quantum mechanics. - It typically is represented by \( h \) and has a value of \( 6.626 \times 10^{-34} \text{ m}^2 \text{kg/s} \).- It relates the energy of a photon to its frequency through the equation: \[ E = h u \] where: - \( E \) = energy, - \( h \) = Planck's constant, and - \( u \) = frequency.Planck's constant is also crucial for finding the wavelengths in the de Broglie wavelength formula:- The equation connects mass, speed, and momentum through Planck's constant to determine a particle's wavelength.- In the exercise solution, Planck's constant was used to find both the electron's and the photon's wavelengths as: \[ \lambda = \frac{h}{p} \]This fundamental constant ties together concepts of particles behaving both like waves and like discrete entities, a core idea in quantum theory.

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Most popular questions from this chapter

(a) Does the de Broglie wavelength of a particle increase or decrease as its kinetic energy increases? Explain. (b) Show that the de Broglie wavelength of an electron in nanometers can be written as \(\lambda=(1.23 \mathrm{nm}) / \sqrt{K},\) where \(K\) is the kinetic energy of the electron in eV. Use classical expressions for momentum and kinetic energy.

The uncertainty in a proton's position is \(0.15 \mathrm{nm}\). (a) What is the minimum uncertainty \(\Delta p\) in its momentum? (b) What is the kinetic energy of a proton whose momentum is equal to this uncertainty \((\Delta p=p) ?\)

Photon A has twice the momentum of photon B. (a) Which photon has the greater wavelength? Explain. (b) If the wave length of photon A is 333 nm, what is the wavelength of photon B?

Light of a particular wavelength does not eject electrons from the surface of a given metal. (a) Should the wavelength of the light be increased or decreased in order to cause electrons to be ejected? (b) Choose the best explanation from among the following: I. The photons have too little energy to eject electrons. To increase their energy, their wavelength should be increased. II. The energy of a photon is proportional to its frequency; that is, inversely proportional to its wavelength. To in crease the energy of the photons so they can eject electrons, one must decrease their wavelength.

An electron and a proton have the same speed. (a) Which has the longer de Broglie wavelength? Explain. (b) Calculate the ratio \(\left(\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}\right)\)

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