Problem 32 - Master Resource Book in JEE Main Physics

Question

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de-Broglie wavelength associated with electron is (use charge of electron \(1.6 \times 10^{-19} \mathrm{C}\), mass of electron \(9.1 \times 10^{-31} \mathrm{~kg}\) \(\left.h=6.623 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (a) \(1.53 \mathrm{~A}\) (b) \(2.53 \mathrm{~A}\) (c) \(3.35 \mathrm{~A}\) (d) \(4.54 \mathrm{~A}\)

Step-by-Step Solution

Answer

The de-Broglie wavelength is approximately \( 1.53 \text{ Å} \), matching option (a).

Step 1: Calculate Kinetic Energy

The kinetic energy (KE) obtained by the electron when accelerated through a potential difference \( V \) is given by \( KE = eV \), where \( e = 1.6 \times 10^{-19} \, \text{C} \) is the electron charge and \( V = 64 \) V. Thus, \[ KE = 1.6 \times 10^{-19} \times 64 \] \[ KE = 1.024 \times 10^{-17} \, \text{J} \].

Step 2: Find the Velocity of the Electron

Use the equation for kinetic energy \( KE = \frac{1}{2}mv^2 \) to find the velocity \( v \) of the electron. Rearrange to find \( v = \sqrt{\frac{2 KE}{m}} \). Substitute \( m = 9.1 \times 10^{-31} \, \text{kg} \) and \( KE = 1.024 \times 10^{-17} \). \[ v = \sqrt{\frac{2 \times 1.024 \times 10^{-17}}{9.1 \times 10^{-31}}} \] \[ v \approx 3.95 \times 10^6 \, \text{m/s} \].

Step 3: Calculate the de Broglie Wavelength

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{mv} \). Substitute the values of \( h = 6.623 \times 10^{-34} \, \text{J}\cdot\text{s} \), \( m = 9.1 \times 10^{-31} \) kg, and \( v = 3.95 \times 10^6 \) m/s: \[ \lambda = \frac{6.623 \times 10^{-34}}{9.1 \times 10^{-31} \times 3.95 \times 10^6} \] \[ \lambda \approx 1.53 \times 10^{-10} \text{ meters} \], or approximately \( 1.53 \text{ Å} \).

Step 4: Compare to Given Options

The calculated wavelength \( 1.53 \text{ Å} \) matches the option (a). Therefore, the correct answer is option (a) \( 1.53 \text{ Å} \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When electrons are accelerated by a potential difference, they gain kinetic energy. This energy can be quantified by the formula:

• \( KE = eV \)

where \( e \) is the charge of an electron \((1.6 \times 10^{-19} \text{ C})\) and \( V \) is the potential difference in volts. In our scenario, the potential difference is 64 volts.
This means the kinetic energy obtained by the electron is \( KE = 1.6 \times 10^{-19} \times 64 = 1.024 \times 10^{-17} \text{ J} \).
Kinetic energy is a fundamental concept affecting how particles move, especially in quantum mechanics.

Potential Difference

Potential difference, often referred to as voltage, is a measure of the energy difference per charge between two points in an electric field. It's what propels electrons, giving them the energy to move through a circuit.
The potential difference drives the electrons, causing them to accelerate. This is crucial in many atomic and subatomic processes.
When an electron is subject to a potential difference of 64 V, it gains energy equal to 64 times its charge. This accumulated energy directly contributes to the kinetic energy of the electron, as seen in the equation \( KE = eV \).
Thus, potential difference transforms into kinetic energy, making electrons useful in electronic devices and quantum experiments.

Electron Velocity

Velocity is a vector quantity that expresses speed with a specific direction. When an electron is accelerated by a potential difference, its velocity increases.
To find the velocity of an electron given its kinetic energy, we use the relation:

• \( KE = \frac{1}{2}mv^2 \)

By rearranging this formula, the velocity (\( v \)) of the electron is derived:

• \( v = \sqrt{\frac{2KE}{m}} \)

Plugging in known values: \( KE = 1.024 \times 10^{-17} \text{ J} \) and \( m = 9.1 \times 10^{-31} \text{ kg} \), we find \( v \approx 3.95 \times 10^6 \text{ m/s} \).
This high velocity indicates how fast quantum particles like electrons move, aiding their role in current flow.

Quantum Mechanics

Quantum mechanics is the branch of physics that deals with phenomena at the nanoscopic scale, particularly in particles like electrons. One key concept here is the de-Broglie wavelength, which describes the wave-like properties of particles.
Louis de Broglie postulated that every moving particle or object of matter has an associated wave. The de-Broglie wavelength \( \lambda \) is given by:

• \( \lambda = \frac{h}{mv} \)

where \( h \) is Planck's constant \(6.623 \times 10^{-34} \text{ J}\cdot\text{s}\), \( m \) is the mass, and \( v \) is the velocity of the electron. Substituting these values, we calculate \( \lambda \approx 1.53 \text{ Å} \).
This result is crucial in explaining the behavior of particles at small scales, influencing how we understand electron interactions in fields like electronics and chemistry.