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Chapter 3: Problem 27

What possibly can be the ratio of the de Broglie wavelengths for two electrons having the same initial energy and accelerated through \(50 \mathrm{~V}\) and \(200 \mathrm{~V}\) ? (a) \(3: 10\) (b) \(10: 3\) (c) \(1: 2\) (d) \(2: 1\)

Short Answer

Expert verified
The ratio is \( 2:1 \), so the correct option is (d).

Step by step solution

01

Understanding de Broglie Wavelength

The de Broglie wavelength \( \lambda \) for a particle is given by the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron, its momentum \( p \) can also be related to its kinetic energy.
02

Relating Kinetic Energy and Momentum

When an electron is accelerated through a potential \( V \), it gains kinetic energy \( eV \). The kinetic energy and momentum \( p \) are related by the formula \( KE = \frac{p^2}{2m} \). By rearranging terms, we get \( p = \sqrt{2m \cdot KE} = \sqrt{2m \cdot eV} \).
03

Expressing de Broglie Wavelength

Substituting \( p = \sqrt{2m \cdot eV} \) into the de Broglie wavelength formula, we have \( \lambda = \frac{h}{\sqrt{2meV}} \). This means the wavelength is inversely proportional to the square root of the potential \( V \), \( \lambda \propto \frac{1}{\sqrt{V}} \).
04

Calculating the Wavelength Ratio

The ratio of the wavelengths for electrons accelerated through \( 50 \text{ V} \) and \( 200 \text{ V} \) is given by \( \frac{\lambda_{50}}{\lambda_{200}} = \frac{\sqrt{200}}{\sqrt{50}} = \sqrt{\frac{200}{50}} = \sqrt{4} = 2 \).
05

Choosing the Correct Option

The computed ratio is \( 2:1 \), which corresponds to option (d). Thus, the ratio of de Broglie wavelengths for the electrons is \( 2: 1 \).

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Key Concepts

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

Kinetic Energy
In physics, kinetic energy is the energy that a body possesses due to its motion. It is defined mathematically as \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
When dealing with electrons, the situation becomes quite interesting, as their kinetic energy changes with acceleration through a potential difference. We often describe this change in energy using the formula \( KE = eV \), where \( e \) is the electron鈥檚 charge and \( V \) is the potential difference they are subjected to.
  • This concept is crucial as it links the electrical potential energy, a form of energy storage, to kinetic energy, manifesting as apparent electron motion.
  • As kinetic energy increases, so does an electron's velocity, which directly affects its momentum and subsequently its de Broglie wavelength.
Understanding how kinetic energy interplays with momentum helps us explain changes in the tiny yet mighty behavior of electrons.
Electron Acceleration
Acceleration occurs when a force is applied to an object, causing a change in motion. When electrons are subjected to an electric field, they experience acceleration. In this context, the electric field is provided by a potential difference.
  • The potential difference works like a driving "push," nudging the electron forward and causing it to zip across the field with increasing velocity.
  • As an electron accelerates, the gain in velocity translates into increased kinetic energy as per the relation \( KE = \frac{1}{2}mv^2 \), emphasizing how acceleration influences energy dynamics.
Understanding electron acceleration is fundamental in electronics and quantum mechanics as it provides insights into how electrons behave under various electric fields, bringing our everyday electronic gadgets to life.
Potential Difference
The potential difference, sometimes described as voltage, is a measure of the work needed to move a charge from one point to another within an electric field. It is expressed in volts (\( V \)) and represents the potential energy per unit charge.
  • For electrons, a higher potential difference means more force exerted upon them, translating to greater acceleration and higher kinetic energy.
  • In our exercise, knowing the potential difference allows us to compute changes in kinetic energy, which in turn affects the de Broglie wavelength of the electron.
By understanding potential difference, we comprehend not only how currents flow in circuits, but also how energy transforms within them. This concept is crucial in both practical applications like circuits and in theoretical explorations like electron behavior in fields.
Momentum
Momentum is a key property that describes the motion of an object. It is calculated as the product of an object's mass and velocity: \( p = mv \). In the context of electrons, managing their momentum is essential to understanding phenomena like diffraction and interference.
  • Due to their high speeds and tiny masses, electrons exhibit behaviors characterizing the dual particle-wave nature of matter.
  • As electrons gain momentum through acceleration, their de Broglie wavelength changes, portraying their wave-like properties.
The de Broglie wavelength formula, \( \lambda = \frac{h}{p} \), establishes a direct relationship between a particle's momentum and its wave characteristics. This understanding furthers our grasp of both classical motion and quantum mechanics, bridging a fundamental connection between the two domains.

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Most popular questions from this chapter

Correct sequence of increasing covalent character is represented oy (a) \(\mathrm{LiCl}, \mathrm{NaCl}, \mathrm{BeCl}_{2}\) (b) \(\mathrm{BeCl}_{2}, \mathrm{NaCl}, \mathrm{LiCl}\) (c) \(\mathrm{NaCl}, \mathrm{LiCl}, \mathrm{BeCl}_{2}\) (d) \(\mathrm{BeCl}_{2}, \mathrm{LiCl}, \mathrm{NaCl}\)

In an atom two electrons are moving in orbits of radii \(r\) and \(9 r\). The ratio of time taken by them to complete one revolution is (a) \(1: 9\) (b) \(1: 3\) (c) \(1: 27\) (d) \(1: 6\)

Which one of the following pairs of species have the same bond order? (a) \(\mathrm{CN}^{-}\)and \(\mathrm{NO}^{+}\) (b) \(\mathrm{CN}\) and \(\mathrm{CN}^{+}\) (c) \(\mathrm{O}_{2}^{-}\)and \(\mathrm{CN}^{-}\) (d) \(\mathrm{NO}^{+}\)and \(\mathrm{CN}^{+}\)

Which hydrogen-like species will have the same radius as that of the Bohr orbit of a hydrogen atom ? (a) \(n=2, \mathrm{Li}^{2+}\) (b) \(n=2, \mathrm{Be}^{3+}\) (c) \(n=2, \mathrm{He}^{+}\) (d) \(n=3, \mathrm{Li}^{2+}\)

The correct order of increasing \(\mathrm{C}-\mathrm{O}\) bond length of \(\mathrm{CO}, \mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}\) is (a) \(\mathrm{CO}_{3}^{2}<\mathrm{CO}_{2}<\mathrm{CO}\) (b) \(\mathrm{CO}_{2}<\mathrm{CO}_{3}^{2-}<\mathrm{CO}\) (c) \(\mathrm{CO}<\mathrm{CO}_{3}^{2-}<\mathrm{CO}_{2}\) (d) \(\mathrm{CO}<\mathrm{CO}_{2}<\mathrm{CO}_{3}^{2-}\)
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