Problem 95 - Chemistry The Central Science In Si Units

Question

An electron is accelerated through an electric potential to a kinetic energy of \(1.6 \times 10^{-15} \mathrm{~J}\). What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is \(E=\frac{1}{2} m v^{2},\) where \(m\) is the mass of the object and \(v\) is the speed of the object.]

Step-by-Step Solution

Answer

The electron's characteristic wavelength is approximately \(3.88 \times 10^{-11} \mathrm{~m}\).

Step 1: Understand the Formula for Wavelength

We need to find the characteristic wavelength of the electron. To do this, we will use the de Broglie wavelength formula: \[\lambda = \frac{h}{p}\]where \(\lambda\) is the wavelength, \(h\) is Planck's constant \((6.626 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} ~ \mathrm{s}^{-1})\), and \(p\) is the momentum of the electron.

Step 2: Determine the Momentum

The momentum \(p\) of an electron can be determined from its velocity and mass. Since the kinetic energy \(E\) of an electron is given, we can relate it to velocity using the equation for kinetic energy:\[E = \frac{1}{2} m v^2\]Rearranging this equation to solve for \(v\), we get:\[v = \sqrt{\frac{2E}{m}} \]where \(m\) is the mass of an electron \((9.109 \times 10^{-31} \mathrm{~kg})\). Inserting the given kinetic energy \(E = 1.6 \times 10^{-15} \mathrm{~J}\), we calculate the velocity. Then, the momentum \(p\) is \(p = mv\).

Step 3: Calculate Velocity

Insert the known values into the velocity equation:\[v = \sqrt{\frac{2 \times 1.6 \times 10^{-15} \mathrm{~J}}{9.109 \times 10^{-31} \mathrm{~kg}}} \]Calculating this, we find \(v \approx 1.875 \times 10^7 \mathrm{~m/s}\).

Step 4: Calculate Momentum

Using the previously calculated velocity, we determine the momentum:\[p = m \times v = 9.109 \times 10^{-31} \mathrm{~kg} \times 1.875 \times 10^7 \mathrm{~m/s} \]\[p \approx 1.708 \times 10^{-23} \mathrm{~kg~m/s} \].

Step 5: Calculate the Wavelength

Finally, substitute the momentum into the de Broglie wavelength formula:\[\lambda = \frac{6.626 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} ~ \mathrm{s}^{-1}}{1.708 \times 10^{-23} \mathrm{~kg~m/s}} \]Calculating, we find \(\lambda \approx 3.88 \times 10^{-11} \mathrm{~m}\).

Key Concepts

Electron Momentum

At the core of understanding the de Broglie wavelength is the concept of momentum, especially for small particles like electrons. Electron momentum is essentially the product of an electron's mass and its velocity. If you're thinking in terms of classical physics, momentum (\(p\) is often expressed as:

• \(p = m imes v\)

This formula involves the electron's mass (\(m\), which is approximately \(9.109 \times 10^{-31} \mathrm{~kg}\), and its calculated velocity (\(v\). Knowing the velocity is crucial because it lets us determine how much momentum the electron has while it moves. It's important to note that as this momentum increases, the wavelength decreases based on the inverse relationship described by the de Broglie equation.
In situations like these, where high speeds and small masses are involved, these values might seem nearly negligible. But in the realm of quantum mechanics, even these tiny differences in momentum make a significant impact on the properties of particles.

Kinetic Energy

Kinetic energy plays a crucial role in determining the velocity of an electron, which in turn helps find its momentum. The kinetic energy (\(E\) of an object is given by the formula:

• \(E = \frac{1}{2} m v^2\)

In our example, the electron is accelerated through a potential to a kinetic energy of \(1.6 \times 10^{-15}\) Joules. This value is used to solve for the velocity of the electron. By rearranging the kinetic energy formula, we have:

• \(v = \sqrt{\frac{2E}{m}}\)

Plugging the known values into this equation lets us find the necessary speed.
This calculated velocity provides vital information to solve for the momentum, as kinetic energy reflects the energy an electron possesses due to its motion, which is vital when utilizing the de Broglie wavelength equation.

Planck's Constant

Planck's constant is a fundamental constant in physics that is central to quantum mechanics and the concept of the de Broglie wavelength. Symbolized as \(h\), it is a proportionality constant with a value approximately equal to \(6.626 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} ~ \mathrm{s}^{-1}\). This constant is crucial when calculating the wavelength from momentum.
In the de Broglie wavelength formula:

• \(\lambda = \frac{h}{p}\)

Planck's constant helps reveal how particles, like electrons, exhibit wave-like properties when in motion. This means that even though particles are small, they still exhibit a wavelength. The beauty of Planck's constant is in showing that at the quantum level, the distinctions between waves and particles become blurred.
Recognizing the role of Planck's constant in the equation helps us appreciate how it bridges the phenomenon of particle-wave duality. This duality is a cornerstone concept in quantum mechanics, emphasizing the intricate nature of microscopic particles and their motions.