Problem 24 - Chemistry and Chemical Reactivity

Question

A beam of electrons \(\left(m=9.11 \times 10^{-31} \mathrm{kg} / \text { electron }\right)\) has an average speed of \(1.3 \times 10^{8} \mathrm{m} / \mathrm{s} .\) What is the wavelength of electrons having this average speed?

Step-by-Step Solution

Answer

The wavelength is \( 5.59 \times 10^{-12} \text{ meters} \).

Step 1: Identify the relevant formula

We need to use the de Broglie wavelength formula to find the wavelength of the electrons. The de Broglie wavelength \( \lambda \) is given by the formula: \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant \( 6.63 \times 10^{-34} \text{Js} \), \( m \) is the mass of an electron, and \( v \) is the velocity of the electron.

Step 2: Substitute the known values into the formula

We substitute the known values into the formula for the de Broglie wavelength. Here, \( m = 9.11 \times 10^{-31} \text{kg} \), and \( v = 1.3 \times 10^8 \text{ m/s} \). Thus:\[ \lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.3 \times 10^8} \]

Step 3: Calculate the wavelength

Perform the calculation to find the wavelength:\[ \lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (1.3 \times 10^8)} = \frac{6.63 \times 10^{-34}}{1.1843 \times 10^{-22}} \approx 5.59 \times 10^{-12} \text{ meters} \]

Step 4: Conclude the result

The wavelength of the electrons having an average speed of \( 1.3 \times 10^8 \text{ m/s} \) is \( 5.59 \times 10^{-12} \text{ meters} \).

Key Concepts

Understanding Electron Beams

An electron beam is essentially a stream of electrons moving in the same direction.
These beams are utilized in various scientific and technical applications, including electron microscopes and cathode ray tubes, like those found in older television sets. The electrons in the beam move at high speeds, which in this context is approximately \(1.3 \times 10^8 \; \mathrm{m/s}\).

When we observe electron beams from a quantum mechanical perspective, they don't just behave as particles—they also exhibit wave-like properties.

• This dual nature is where the concept of de Broglie wavelength comes into play.
• Understanding this wave property is crucial for explaining phenomena such as interference patterns in electron diffraction.

In our example exercise, by calculating the de Broglie wavelength, we get insight into the wave aspect of electrons at a high velocity. This understanding is vital in fields like quantum mechanics and particle physics, giving us a peek into the nature of very small particles.

The Role of Planck's Constant

Planck's constant is a fundamental constant in physics, symbolized by \(h\), and its value is \(6.63 \times 10^{-34} \; \mathrm{Js}\).
This constant plays a significant role in various quantum mechanics formulas, including the de Broglie wavelength calculation.

Planck's constant quantifies the size of discrete units, or 'quanta,' of energy.

• In the de Broglie equation, Planck's constant connects the particle-like and wave-like behaviors of electrons.
• It allows us to calculate how the mass and speed of electrons translate to a measurable wavelength.

This is crucial because it enables us to predict and understand the behavior of particles at atomic and subatomic levels.
With Planck's constant, we are equipped with the ability to measure wave characteristics of particles, thereby expanding our comprehension of the physical universe.

Wavelength Calculation for Electron Beams

Calculating the wavelength of an electron beam involves using the de Broglie wavelength formula, given by: \[\lambda = \frac{h}{mv}\]
Here, \(\lambda\) represents the wavelength, \(h\) is Planck's constant, \(m\) is the mass of an electron, and \(v\) is the velocity of the electron.

To calculate the wavelength:

• Substitute the known values: \(h = 6.63 \times 10^{-34} \; \mathrm{Js}\), \(m = 9.11 \times 10^{-31} \; \mathrm{kg}\), and \(v = 1.3 \times 10^8 \; \mathrm{m/s}\).
• Perform the division: \[ \lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (1.3 \times 10^8)} \]

This results in a wavelength of approximately \(5.59 \times 10^{-12}\) meters, showing the wave nature of electrons.
Understanding how to perform this calculation is essential for exploring the quantum qualities of matter, which have applications ranging from electron microscopy to quantum computing, providing insights into how matter behaves at a foundational level.