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

Short Answer

Expert verified
By using the calculated wavelength of the electron and the given mass of a proton, the speed of a proton having the same wavelength as the electron can be calculated.

Step by step solution

01

Understanding de Broglie's Wavelength

De Broglie wavelength describes the wave nature of particles. According to this concept, a particle of momentum p (which is the product of mass and velocity m*v) is associated with a wave of wavelength \(\lambda\), which is given by de Broglie's equation, \(\lambda = \frac{h}{m*v}\) where \(h\) represents Planck’s constant.
02

Calculating the de Broglie Wavelength of the Electron

Given that the electron is moving at a speed of \(8.00 \times 10^{6}\) m/s, the mass of the electron \(m_e\) is \(9.11 \times 10^{-31}\) kg, and \(h\) (Planck’s constant) is \(6.63 \times 10^{-34}\) Js, using the de Broglie formula, we can calculate the wavelength of the electron as \(\lambda_e = \frac{h}{m_e* v_e}\). Substitute the given values into the formula to get the value of \(\lambda_e\).
03

Finding the Speed of the Proton with the same Wavelength

For a proton and electron to have the same de Broglie wavelength, we can equate their de Broglie formulas. This gives \(\frac{h}{m_e* v_e} = \frac{h}{m_p* v_p}\), which simplifies to \(m_e* v_e = m_p* v_p\). Substituting for \(m_e\), \(v_e\) and \(m_p\) (mass of proton), we can solve for \(v_p\) (speed of the proton). The mass of a proton \(m_p\) is \(1.67 \times 10^{-27}\) kg.

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Key Concepts

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

Wave-Particle Duality
One of the most intriguing concepts in quantum mechanics is wave-particle duality, a fundamental principle that tells us particles at the quantum level exhibit both wave-like and particle-like properties. This dual nature is a cornerstone of quantum mechanics and was first postulated by French physicist Louis de Broglie in 1924. De Broglie proposed that particles such as electrons, which were traditionally considered to just have mass and move through space like tiny billiard balls, also have an associated wavelength.

Imagine tossing a pebble into a pond. The pebble, while clearly a particle, creates ripples spreading out in a wave pattern. In a similar way, electrons, when in motion, behave as if they are surrounded by these ripples—these are their de Broglie wavelengths. This concept is especially significant because it leads to the idea that particles can interfere and diffract, much like waves—a property that is observable in phenomena such as the famous double-slit experiment.
Momentum
Momentum is a key concept when discussing the movement of objects, both in classic physics and quantum mechanics. It's defined as the product of an object's mass and velocity. Mathematically, it is represented as p = m * v, where p stands for momentum, m is the mass, and v is the velocity of the object.

In the context of wave-particle duality, the momentum of a particle is directly related to its de Broglie wavelength. The de Broglie relation \( \lambda = \frac{h}{p} \) connects the momentum of a particle with its wavelength, meaning that as the momentum of a particle increases, its de Broglie wavelength decreases, and vice versa. In quantum terms, understanding momentum is crucial, because it helps predict the behavior of particles at a subatomic level, and forms the bedrock for various quantum mechanical models and theories.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It challenges and extends classical mechanics, which fails to describe such phenomena adequately. Quantum mechanics introduces the concept that energy is quantized and adopts a probabilistic approach to predict the behavior of particles.

In quantum mechanics, the de Broglie wavelength becomes crucial when examining and predicting the location and momentum of particles—which is inherently uncertain. The famous Heisenberg uncertainty principle directly arises from this wave-particle duality, stating that one cannot simultaneously know both the exact position and exact momentum of a particle. This introduces an inherent fuzziness in quantum predictions—a stark contrast to the determinism that pervades classical mechanics.
Planck's Constant
Planck's constant, denoted h, is a physical constant that plays a vital role in quantum mechanics. Its value is approximately \(6.626 \times 10^{-34}\) joule seconds. Planck's constant is a scaling factor between the microscopic world of quantum mechanics and the macroscopic world we observe. It sets the scale of quantum effects and relates energy and frequency through the equation E = h * ν, where E is the energy of a quantum of electromagnetic radiation, and ν is the frequency.

In the context of de Broglie's hypothesis, Planck's constant links the momentum of particles to their wave-like characteristics, as we've seen in the equation \(\lambda = \frac{h}{m*v}\). Planck's constant is so fundamental to quantum mechanics that it appears in the Heisenberg uncertainty principle and the Schrödinger equation, which underpins the whole framework of quantum physics.

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Most popular questions from this chapter

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda=h / p\) still holds, but we must use the relativistic expression for momentum, \(p=m v / \sqrt{1-v^{2} / c^{2}}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is \(v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}\) (b) The quantity \(h / m c\) equals \(2.426 \times 10^{-12} \mathrm{~m}\). (As we saw in Section 38.3 , this same quantity appears in Eq. \((38.7),\) the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h / m c,\) the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to \(c .\) In this case it is convenient to write \(v=(1-\Delta) c\) and express the speed of the electron in terms of \(\Delta\) rather than \(v .\) Find an expression for \(\Delta\) valid when \(\lambda \ll h / m c .[\)Hint: Use the binomial expansion \((1+z)^{n}=1+n z+\left[n(n-1) z^{2} / 2\right]+\cdots,\) valid for the case \(|z|<1 .]\) (c) How fast must an electron move for its de Broglie wavelength to be \(1.00 \times 10^{-15} \mathrm{~m}\), comparable to the size of a proton? Express your answer in the form \(v=(1-\Delta) c\), and state the value of \(\Delta\).

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n=3\) level. Determine the wavelength and frequency of the photon.

A hydrogen atom is in a state with energy \(-1.51 \mathrm{eV}\). In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Why is it easier to use helium ions rather than neutral helium atoms in such a microscope? (a) Helium atoms are not electrically charged, and only electrically charged particles have wave properties. (b) Helium atoms form molecules, which are too large to have wave properties. (c) Neutral helium atoms are more difficult to focus with electric and magnetic fields. (d) Helium atoms have much larger mass than helium ions do and thus are more difficult to accelerate.

(a) What accelerating potential is needed to produce electrons of wavelength \(5.00 \mathrm{nm} ?\) (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

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