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 \(\mathrm{V}\) 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) meV

Step-by-Step Solution

Answer

(c) \( \sqrt{2meV} \)

Step 1: Understand the problem

We need to find the value of \( \frac{h}{\lambda} \) for electrons accelerated by a potential difference \( V \). We use the concepts of electron dynamics and De Broglie's wavelength.

Step 2: Relation between velocity and potential difference

When an electron is accelerated by a potential difference \( V \), its kinetic energy \( KE \) is equal to \( eV \), where \( e \) is the charge of the electron. Also, \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the electron. Equating both, we get \( \frac{1}{2}mv^2 = eV \), leading to \( v = \sqrt{\frac{2eV}{m}} \).

Step 3: Apply De Broglie's hypothesis

De Broglie's wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \), where \( p \) is the linear momentum. The linear momentum \( p \) can be expressed as \( mv \). Plug in the velocity from Step 2 to get \( p = m \cdot \sqrt{\frac{2eV}{m}} \) simplifying to \( \sqrt{2meV} \).

Step 4: Calculate the value of \( \frac{h}{\lambda} \)

From \( \lambda = \frac{h}{p} \) and \( p = \sqrt{2meV} \), we calculate \( \frac{h}{\lambda} = p = \sqrt{2meV} \).

Step 5: Verify against options

The derived expression \( \sqrt{2meV} \) matches option (c).

Key Concepts

De Broglie Wavelength

De Broglie wavelength is a fundamental concept in quantum mechanics that describes the wave-like nature of particles. According to Louis de Broglie's hypothesis, all matter exhibits wave-like properties, and the wavelength associated with a particle is known as its De Broglie wavelength. It is expressed by the equation: \[ \lambda = \frac{h}{p} \] where

• \( \lambda \) is the De Broglie wavelength,
• \( h \) is Planck’s constant, and
• \( p \) is the momentum of the particle.

For an electron moving with velocity \( v \), the momentum \( p \) can be given as \( mv \), where \( m \) is the mass of the electron. Thus, the equation becomes: \[ \lambda = \frac{h}{mv} \] This indicates that the wavelength is inversely proportional to the momentum of the electron. This relationship helps us understand how subatomic particles, like electrons, can behave like waves under certain conditions. It is a striking contrast to classical physics, where such wave-particle duality is not observed.

Kinetic Energy

Kinetic energy is the energy that a particle possesses due to its motion. For an electron, kinetic energy can be calculated based on its velocity. The formula for kinetic energy \( KE \) is: \[ KE = \frac{1}{2}mv^2 \] where

• \( m \) is the mass of the electron, and
• \( v \) is its velocity.

In the context of this exercise, the kinetic energy of an electron is derived from the potential difference \( V \) through which it has been accelerated. Thus, \( KE = eV \), where \( e \) is the charge of the electron and \( V \) is the potential difference. Equating \( \frac{1}{2}mv^2 = eV \), we can solve for velocity, which links kinetic energy directly with the motion imparted by the electrical potential. This connection is vital for calculating the momentum used in determining the De Broglie wavelength.

Potential Difference

A potential difference, often measured in volts, refers to the difference in electric potential energy per unit charge between two points. For electrons, this concept is crucial as it describes how much energy is provided to or taken from the electron as it moves through an electric field. When an electron is accelerated by a potential difference \( V \), it gains kinetic energy stated by the equation: \[ KE = eV \] Here,

• \( e \) is the charge of the electron,
• \( V \) is the potential difference in volts.

The relationship between potential difference and kinetic energy explains how electrons gain speed and thus momentum when subjected to a voltage. This gained kinetic energy is fundamental in describing electron velocity and consequently their De Broglie wavelength, thus influencing their behavior in quantum experiments and applications.

Electron Velocity

Electron velocity is a key element in understanding electron dynamics as it affects both the kinetic energy and the momentum, thereby influencing the De Broglie wavelength. When an electron is subject to a potential difference \( V \), the velocity \( v \) of the electron can be approximated using the derived formula: \[ v = \sqrt{\frac{2eV}{m}} \] Here,

• \( e \) is the charge of the electron,
• \( V \) is the potential difference, and
• \( m \) is the mass.

This formula arises from equating the kinetic energy (\( KE = \frac{1}{2}mv^2 \)) with the energy provided by the potential difference (\( eV \)). By solving for \( v \), one can predict how fast an electron is moving when accelerated by its environment. This velocity calculation is essential for understanding how electrons behave in devices such as cathode ray tubes or electron microscopes, where precise control over velocity is required.