An electron and a photon each have a wavelength of 0.20 nm . What is the momentum (in kg ·m / s

step 1 For part A, we know that the momentum of the photon, P, is going to be equal to Planks constant,

An electron and a photon each have a wavelength of 0.20 nm ....

An electron and a photon each have a wavelength of 0.20 $\mathrm{nm} .$ What is the momentum (in $\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} )$ of the (a) electron and
(b) photon? What is the energy (in eV) of the (c) electron and
(d) photon?

Key Concepts

De Broglie Hypothesis

This concept states that every particle has an associated wavelength, given by ? = h/p, where h is Planck's constant and p is the momentum. This relationship applies to all matter and is foundational in quantum mechanics, linking the wave nature to particles such as electrons.

Photon Energy-Momentum Relationship

Photons, which are massless particles, obey the relation E = pc, where E is energy, p is momentum, and c is the speed of light. Additionally, since the wavelength of a photon is related to its momentum by p = h/?, its energy can also be expressed as E = hc/?. This concept is crucial for understanding the behavior and energy quantization of light.

Classical Kinetic Energy for Non-relativistic Particles

For particles like electrons moving at speeds much less than the speed of light, the kinetic energy can be calculated using the expression K = p²/(2m), where m is the particle's mass. This formula, derived from classical mechanics, is applicable when relativistic effects are negligible.

Energy Unit Conversion (Joules to Electronvolts)

Energy is often expressed in joules in SI units, but in atomic and particle physics, it is common to use electronvolts (eV) for convenience. The conversion factor between these units is 1 eV = 1.602 x 10?¹? joules. This conversion is essential when comparing energies computed in joules to typical energy scales in quantum systems.

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