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How does the wavelength of a helium ion compare to that of an electron accelerated through the same potential difference? (a) The helium ion has a longer wavelength, because it has greater mass. (b) The helium ion has a shorter wavelength, because it has greater mass. (c) The wavelengths are the same, because the kinetic energy is the same. (d) The wavelengths are the same, because the electric charge is the same.

Short Answer

Expert verified
The helium ion has a shorter wavelength, because it has greater mass.

Step by step solution

01

Understand the de Broglie wavelength equation

The de Broglie wavelength equation is \( \lambda = h / p \) where \( h \) is the Planck's constant and \( p \) is the momentum. The smaller the momentum, the larger the wavelength.
02

Determine the momentum

The momentum of a particle is given by the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. The velocity \( v \) of the particle is determined by the kinetic energy, which is the same for the helium ion and the electron because they are accelerated through the same potential difference. Therefore, the momentum of the helium ion is larger because it has a larger mass.
03

Compare the wavelengths

Since the helium ion has a larger momentum due to its larger mass, according to the de Broglie wavelength equation, it will have a smaller wavelength. If we algebraically rearrange the de Broglie wavelength equation, a direct relationship between mass and wavelength is seen: the greater the mass, the smaller the wavelength and vice versa.

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Key Concepts

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

Momentum in Physics
Understanding momentum is crucial when addressing problems in physics, especially in quantum mechanics and the behavior of particles at the atomic level. Momentum, represented by the symbol 'p', is the product of the mass (m) of an object and its velocity (v), denoted by the equation:
\[p = mv\].
Momentum is a vector quantity, meaning it has both magnitude and direction. It's a measure of the 'quantity of motion' that an object possesses and plays a significant role in the analysis of collisions and movements.

In the context of quantum mechanics, the momentum of a particle is linked to its de Broglie wavelength; they are inversely proportional. This means that as the momentum of a particle increases, its de Broglie wavelength decreases, and vice versa. This relationship is central to understanding the dual nature of matter, where particles exhibit both wave-like and particle-like properties.

When two particles are accelerated through the same potential difference, they gain the same kinetic energy. However, their momenta will differ if they have different masses. A greater mass results in greater momentum when velocity is kept constant. Consequently, this affects the associated de Broglie wavelength, solidifying the intimate connection between momentum and wave-like behavior in the quantum realm.
Particle-Wave Duality
At the heart of quantum mechanics lies the notion of particle-wave duality, a principle suggesting that every particle or quantic entity may be described as either a particle or a wave. This duality is best understood through the lens of the de Broglie hypothesis, which states that matter particles, such as electrons or helium ions, exhibit wave-like characteristics under the right conditions.

The de Broglie wavelength of a particle is given by the equation: \[ \text{{de Broglie wavelength}} (\lambda) = \frac{{h}}{{p}} \], where 'h' is Planck's constant—a fundamental constant in quantum mechanics—and 'p' is the momentum of the particle. This relationship reveals that the wave nature of particles is more evident when they have lower momentum; high-speed, low-mass particles have longer wavelengths, making their wave properties more observable.

This concept is fundamental when analyzing the behavior of particles in different scenarios, such as electrons moving through electric circuits or the diffraction patterns created when particles travel through a slit. Understanding the de Broglie wavelength helps explain otherwise mysterious phenomena, like the appearance of interference patterns that can only be produced by waves, even though they are brought about by particles.
Potential Difference
In the field of physics, specifically electromagnetism, potential difference is a core concept that describes the work needed to move a charge from one point to another in an electric field. It's represented by the unit 'volt' and often referred to as voltage.

A potential difference is applied to particles in many experimental setups to accelerate them, which imparts kinetic energy. The kinetic energy (KE) acquired by a charged particle that has been accelerated through a potential difference (V) is given by the equation: \[ KE = eV \], where 'e' is the elementary charge of the particle. This relationship indicates that regardless of the mass, two particles with the same charge will gain the same kinetic energy when accelerated through the same potential difference.

However, it's important to remember that while the kinetic energy might be the same, the velocity will differ for particles of different masses. Helium ions, for example, being much heavier than electrons, will have a smaller velocity after being accelerated through the same potential difference. This variance in velocity has a direct impact on the momentum and in turn, as per the de Broglie equation, affects the wavelength of the particles. Thus, while operating under the same potential difference, distinct masses yield distinct momenta, leading to a difference in de Broglie wavelengths for different particles.

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Most popular questions from this chapter

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 \(6.50 \times 10^{-14} \mathrm{~m}\) 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 \(6.64 \times 10^{-27} \mathrm{~kg} .\) (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in \(\mathrm{MeV}\). (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

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 beam of electrons is accelerated from rest through a potential difference of \(0.100 \mathrm{kV}\) and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at \(\pm 14.6^{\circ}\) from the original direction of the beam. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

CP Electrons go through a single slit \(300 \mathrm{nm}\) wide and strike a screen \(24.0 \mathrm{~cm}\) away. At angles of \(\pm 20.0^{\circ}\) from the center of the diffraction pattern, no electrons hit the screen, but electrons hit at all points closer to the center. (a) How fast were these electrons moving when they went through the slit? (b) What will be the next pair of larger angles at which no electrons hit the screen?

A hydrogen atom in an excited bound state labeled with principal quantum number \(n=3\) absorbs a photon that has wavelength \(\lambda\). The atom is ionized and the electron has kinetic energy \(8.00 \mathrm{eV}\) after it has left the atom. What was the wavelength \(\lambda\) of the photon?

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