Problem 42 - College Physics

Question

Find the wavelengths of a photon and an electron that have the same energy of \(25 \mathrm{eV}\). (The energy of the electron is its kinetic energy.)

Step-by-Step Solution

Answer

The photon wavelength is approximately \(4.97 \times 10^{-8}\) m, and the electron wavelength is approximately \(7.00 \times 10^{-10}\) m.

Step 1: Convert energy to joules

First, we need to convert the given energy of 25 eV (electron volts) into joules because the standard SI unit for energy calculations in physics is joules. We use the conversion factor: 1 eV = 1.602 x 10^(-19) J. Thus, 25 eV = 25 * 1.602 x 10^(-19) J = 4.005 x 10^(-18) J.

Step 2: Calculate the wavelength of a photon

We use the formula for the energy of a photon: E = \( \frac{hc}{\lambda} \),where h = 6.626 x 10^(-34) J·s (Planck's constant) and c = 3.0 x 10^8 m/s (speed of light).Rearrange to find \( \lambda \):\( \lambda = \frac{hc}{E} \).Substitute the values:\( \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{4.005 \times 10^{-18}} \approx 4.97 \times 10^{-8} \) m.

Step 3: Calculate the wavelength of an electron

The de Broglie wavelength formula for particles is:\( \lambda = \frac{h}{p} \),where \( p \) is the momentum. For an electron, energy is kinetic, given by:E = \( \frac{1}{2}mv^2 \). However, we also know \( E = \frac{p^2}{2m} \).Thus, \( p = \sqrt{2mE} \), where m = 9.109 x 10^(-31) kg (mass of an electron).Calculate p:\( p = \sqrt{2 \times 9.109 \times 10^{-31} \times 4.005 \times 10^{-18}} \approx 9.48 \times 10^{-25} \) kg·m/s.Finally, find \( \lambda \):\( \lambda = \frac{6.626 \times 10^{-34}}{9.48 \times 10^{-25}} \approx 7.00 \times 10^{-10} \) m.

Key Concepts

de Broglie Wavelength

The concept of de Broglie wavelength beautifully merges the ideas of waves and particles. It's a cornerstone of quantum mechanics, explaining that every particle has a wave associated with it. This wave's wavelength, known as the de Broglie wavelength, depends on the particle's momentum. Momentum is mathematically represented as the product of mass and velocity. As expressed by de Broglie,

• For any particle, the wavelength, \( \lambda \), is given by: \( \lambda = \frac{h}{p} \).
• Here, \( h \) is Planck's constant, a fundamental quantity in quantum mechanics.
• \( p \) represents the momentum of the particle.

To find an electron's de Broglie wavelength, first calculate its momentum. For an electron with kinetic energy \( E \), we have: \( p = \sqrt{2mE} \). This means that to compute the wavelength, you replace the momentum in the formula with this expression. Doing so allows you to find how, even at rest, molecules have this wave behavior whenever they move, reflecting their dual wave-particle nature.

Photon Energy

In the quantum world, photons are unique. They do not have mass like electrons, yet they carry energy. This energy is directly associated with their wave-like nature and is inversely related to their wavelength. The relationship is captured in the formula:

• \( E = \frac{hc}{\lambda} \), where \( E \) denotes photon energy, and \( \lambda \) is the wavelength.
• \( h \) is Planck's constant, and \( c \) is the speed of light.

This formula reveals that as the wavelength decreases, the energy increases, highlighting the compact, high-energy nature of short-wavelength photons. When considering a photon with a known energy, such as 25 electron volts (eV), you can rearrange the formula to find its wavelength: \( \lambda = \frac{hc}{E} \). By using this inverse relationship, you find that higher energy results in shorter wavelengths. This insight into photons helps bridge classical physics with modern quantum mechanics, revealing the particle's nature through energy.

Electron Kinetic Energy

In classical physics, objects in motion have kinetic energy, proportional to their mass and speed. However, electrons in the quantum realm possess this energy in a highly wave-oriented aspect. The classical formula for kinetic energy is:

• \( E_k = \frac{1}{2} mv^2 \), where \( E_k \) is the kinetic energy.
• \( m \) is the mass of the electron; \( v \) is its velocity.

However, for a more practical calculation, especially at high velocities, the energy-momentum relationship in quantum mechanics is often used: \( E = \frac{p^2}{2m} \). This shows how momentum and energy relate, making it possible to calculate the electron's speed through its kinetic energy. Kinetic energy thus not only speaks to the motion but also to the wave behavior of electrons, highlighting quantum mechanics' dual nature of particles. By measuring kinetic energy, we get insight into both particles' speed and their wavelength, underscoring these particles' fascinating wave-particle duality.