Problem 181 - A Complete Resource Book in Chemistry for JEE Main

Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and \(\mathrm{m}\) are charge and mass of an electron, respectively, then the value of \(\mathrm{h} / \lambda\) (where \(\lambda\) is wavelength associated with electron wave) is given by: [2016] (a) \(2 \mathrm{meV}\) (b) \(\sqrt{\mathrm{meV}}\) (c) \(\sqrt{2 \mathrm{meV}}\) (d) \(\mathrm{meV}\)

Step-by-Step Solution

Answer

The correct answer is \(\sqrt{2\mathrm{meV}}\) (option c).

Step 1: Identify the Known Values

We are given that the electrons are accelerated through a potential difference of \( V \) esu, and the charge \( e \) and mass \( m \) of an electron are known.

Step 2: Relate Kinetic Energy and Potential Difference

The kinetic energy \( K \) gained by an electron when accelerated through potential \( V \) is given by:\[ K = eV \]

Step 3: Relate De Broglie Wavelength to Momentum

For an electron with momentum \( p \), the De Broglie wavelength \( \lambda \) is given by:\[ \lambda = \frac{h}{p} \]Hence, \( \frac{h}{\lambda} = p \).

Step 4: Express Momentum in Terms of Kinetic Energy

The momentum \( p \) of an electron can also be expressed as:\[ p = \sqrt{2mK} \]Substituting \( K = eV \), we have:\[ p = \sqrt{2meV} \]

Step 5: Solve for \( \frac{h}{\lambda} \)

From Step 3, we know that \( \frac{h}{\lambda} = p \), and substituting \( p \) from Step 4, we get:\[ \frac{h}{\lambda} = \sqrt{2meV} \]

Step 6: Match with Given Options

Comparing \( \frac{h}{\lambda} = \sqrt{2meV} \) with the options given, option (c) is correct:(c) \(\sqrt{2meV}\).

Key Concepts

Electron Acceleration

Electron acceleration is a pivotal concept in understanding how an electron gains energy and velocity. When a charged particle like an electron is placed in an electric field induced by a potential difference, it accelerates. This potential difference, represented as \( V \), is key in determining how fast and energetic the electron becomes. As electrons accelerate through this potential difference, they gain kinetic energy, which is directly related to the strength of the electric field provided by the potential difference.

Key aspects to remember about electron acceleration:

• The potential difference between two points causes the electron to accelerate.
• The greater the potential difference, the more accelerated the electron becomes.
• Acceleration results from the force exerted by the electric field on the electron's charge \( e \).

Knowing how acceleration works helps us further our understanding of electron behaviors in electric fields.

Kinetic Energy and Potential Difference

Kinetic energy and potential difference are directly connected when it comes to analyzing electrons moving in an electric field. When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equivalent to \( eV \). This relationship is critical to determining various properties of the electron's motion.

This relationship can be broken down as follows:

• Kinetic energy \( K \) gained by an electron is calculated using \( K = eV \).
• \( e \) is the charge of an electron, and \( V \) is the potential difference the electron passes through.
• This equation shows how voltage directly impacts the kinetic energy of charged particles.

By understanding this connection, we get insights into how energy from electric fields is converted into the movement and momentum of electrons.

Momentum and Wavelength Relationship

The relationship between momentum and wavelength is a fundamental aspect of quantum mechanics, especially when discussing particles like electrons. This relationship is best described by the De Broglie wavelength, which links the wave nature of particles like electrons with their momentum.

Here's how to make sense of this relationship:

• De Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum.
• Momentum \( p \) of an electron can also be expressed in terms of kinetic energy as \( p = \sqrt{2mK} \).
• Substituting kinetic energy \( K = eV \), we find that \( p = \sqrt{2meV} \).

Thus, the momentum and wavelength of an electron are linked in a way that when momentum changes, so does the wavelength. Understanding this equivalence is crucial for interpreting the wave-particle duality of matter, crucial for analyzing electron behaviors at microscopic scales.