Ch 39: Particles Behaving as Waves

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 39: Particles Behaving as Waves

Problem 5

Chapter 38, Problem 5

An electron is moving with a speed of $8.00 \times 10^{6}$ m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

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The de Broglie wavelength (λ) is given by the formula:

λ = \frac{h}{p}

, where

h

is Planck's constant and

p

is the momentum of the particle.

The momentum

p

is related to the mass and velocity of the particle by the formula:

p = m v

, where

m

is the mass and

v

is the velocity.

Since the de Broglie wavelength is the same for both the electron and the proton, their momenta must also be equal. Set the momentum of the electron equal to the momentum of the proton:

m_{e} v_{e} = m_{p} v_{p}

, where

m_{e}

and

m_{p}

are the masses of the electron and proton, respectively, and

v_{e}

and

v_{p}

are their velocities.

Rearrange the equation to solve for the velocity of the proton:

v_{p} = \frac{m_{e}}{v_{e}}

. Substitute the known values: the mass of the electron

m_{e} = 9.11 \times 10^- 31

kg, the mass of the proton

m_{p} = 1.67 \times 10^- 27

kg, and the velocity of the electron

v_{e} = 8.00 \times 10^6

m/s.

Perform the substitution and simplify the expression to find the velocity of the proton. Note that the proton's velocity will be much smaller than the electron's velocity due to the proton's significantly larger mass.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that relates the wavelength of a particle to its momentum. It is given by the formula λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. This concept illustrates the wave-particle duality of matter, indicating that particles like electrons and protons exhibit both wave-like and particle-like properties.

Momentum

Momentum is a vector quantity defined as the product of an object's mass and its velocity (p = mv). In the context of the de Broglie wavelength, momentum plays a crucial role as it directly influences the wavelength of a particle. For an electron and a proton to have the same de Broglie wavelength, their momenta must be equal, which requires understanding their respective masses and velocities.

Mass of Particles

The mass of a particle is a measure of its resistance to acceleration when a force is applied. In this context, the electron has a much smaller mass (approximately 9.11 x 10^-31 kg) compared to the proton (approximately 1.67 x 10^-27 kg). This difference in mass is significant when calculating the speed of the proton that would yield the same de Broglie wavelength as the electron, as it affects the relationship between mass, velocity, and momentum.

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Related Practice

Textbook Question

Through what potential difference must electrons be accelerated if they are to have:

(a) the same wavelength as an x ray of wavelength $0.220$ nm; and

(b) the same energy as the x ray in part (a)?

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Textbook Question

An electron has a de Broglie wavelength of $2.80 \times 10^{- 10}$ m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

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Textbook Question

Calculate the de Broglie wavelength of a $5.00$ -g bullet that is moving at $340$ m/s. Will the bullet exhibit wavelike properties?

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Textbook Question

For crystal diffraction experiments (discussed in Section $39.1$ ), wavelengths on the order of $0.20$ nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is a photon.

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Textbook Question

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in 'head-on' to a particular lead nucleus and stops $6.50 \times 10^{- 14}$ m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has $82$ protons, remains at rest. The mass of the alpha particle is $6.64 \times 10^{- 27}$ kg.

(a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV.

(b) What initial kinetic energy (in joules and in MeV) did the alpha particle have?

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Textbook Question

An alpha particle ( $m = 6.64 \times 10^{- 27}$ kg) emitted in the radioactive decay of uranium- $238$ has an energy of $4.20$ MeV. What is its de Broglie wavelength?

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An electron is moving with a speed of 8.00×1068.00\(\times\)10^6 m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

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Key Concepts

de Broglie Wavelength

Momentum

Mass of Particles

An electron is moving with a speed of $8.00 \times 10^{6}$ m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?