Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

Particle A moving with a certain velocity has de Broglie wavelength of \(1 \AA\). If the particle B has mass \(20 \%\) and velocity \(80 \%\) of that of \(A\), the de Broglie wavelength of \(\mathrm{B}\) will be

Step-by-Step Solution

Answer

The de Broglie wavelength of particle B is \( 6.25 \text{ Å}.\)

Step 1: Understanding de Broglie Wavelength

The de Broglie wavelength \( \lambda \) for a particle is given by: \( \lambda = \frac{h}{mv} \) where \( h \) is Planck's constant, \( m \) is the mass, and \( v \) is the velocity of the particle.

Step 2: de Broglie Wavelength of Particle A

For particle A, we are given its de Broglie wavelength as \( \lambda_A = 1 \text{ Å} \). This implies \( \lambda_A = \frac{h}{m_A v_A} = 1 \times 10^{-10} \text{ m}. \) Here, \( m_A \) and \( v_A \) are the mass and velocity of particle A.

Step 3: Defining Mass and Velocity of Particle B

We know particle B has mass \( m_B = 0.2 m_A \) and velocity \( v_B = 0.8 v_A \).

Step 4: Calculating de Broglie Wavelength of Particle B

The de Broglie wavelength for particle B is given by: \( \lambda_B = \frac{h}{m_B v_B} = \frac{h}{0.2 m_A \times 0.8 v_A} = \frac{h}{0.16 m_A v_A}. \) Since \( \lambda_A = \frac{h}{m_A v_A} = 1 \times 10^{-10} \text{ m}, \lambda_B = \frac{1}{0.16} \lambda_A.\)

Step 5: Simplifying and Finding the Wavelength of B

Calculate \( \frac{1}{0.16} \lambda_A = 6.25 \times 1 \times 10^{-10} \text{ m} = 6.25 \text{ Å}. \) Therefore, the de Broglie wavelength of particle B is \( 6.25 \text{ Å}. \)

Key Concepts

Planck's Constant

Planck's constant, often denoted by the symbol \( h \), is a fundamental constant in quantum physics. Its approximate value is \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \). This constant is crucial for calculating various quantum phenomena, including the de Broglie wavelength.
It correlates the frequency of a photon to its energy, illustrating the wave-particle duality of matter. In the context of de Broglie's hypothesis, it helps us understand how particles exhibit wave-like properties. De Broglie used Planck's constant to formulate an equation that relates the properties of mass, velocity, and wavelength of a particle.

• Planck's constant is a universal constant with broad applications.
• It bridges the relationship between observable quantum effects and classical mechanics.
• In the equation for de Broglie wavelength \( \lambda = \frac{h}{mv} \), it is the numerator, highlighting its significance in wave calculations.

Without Planck's constant, our understanding of the quantum realm would be incomplete.

Mass and Velocity Relationship

The relationship between mass and velocity is essential when analyzing particle dynamics. In classical mechanics, both variables are part of the basic equation for momentum \( p = mv \). In the quantum context of de Broglie's hypothesis, these factors determine the wavelength of a particle.
When you decrease the mass or velocity of a particle, its momentum decreases, leading to a longer wavelength. In the exercise, if Particle B's mass is \( 20\% \) of Particle A's and its velocity is \( 80\% \) of Particle A's velocity, the combined effect on momentum is significant.

• For Particle B, the mass \( m_B = 0.2 \, m_A \).
• The velocity \( v_B = 0.8 \, v_A \).
• Momentum adjustment: \( m_B v_B = 0.2 \, m_A \times 0.8 \, v_A = 0.16 \, m_A v_A \).

This combined change in mass and velocity causes the particle to have a markedly different de Broglie wavelength compared to Particle A.

Wavelength Calculation

Calculating the wavelength of particles using de Broglie's equation helps us quantify and compare their wave properties. The equation \( \lambda = \frac{h}{mv} \) provides a mathematical description where the wavelength \( \lambda \) is inversely related to the momentum.
For Particle A, we know its de Broglie wavelength is \( 1 \text{ Å} \). For Particle B, with given mass and velocity, we can determine its wavelength. Using the adjusted mass and velocity:

• Given \( m_B = 0.2 \, m_A \) and \( v_B = 0.8 \, v_A \), plug into the formula:
• \( \lambda_B = \frac{h}{0.16 \, m_A v_A} \).
• Recognizing \( \lambda_A = \frac{h}{m_A v_A} = 1 \text{ Å} \), calculate \( \lambda_B = \frac{1}{0.16} \lambda_A \).
• Simplifying gives \( \lambda_B = 6.25 \text{ Å} \).

This calculation showcases how changes in mass and velocity significantly influence the behavior and properties of quantum particles. Understanding this helps elucidate the nature of matter and light on a fundamental level.