Question

A photon has the same momentum as an electron moving at \(10^{6} \mathrm{~m} / \mathrm{s}\). (a) Determine the photon's wavelength \(\mathrm{N}\). (b) What is the ratio of the kinetic energies of the two?

Short answer

The wavelength of the photon, when its momentum equals that of an electron moving at \(10^6\ m/s\), is approximately \(7.3 \times 10^{-7}\ m\). The ratio of the kinetic energies of the photon to the electron, given these parameters, is approximately 1836.

Step-by-step solution

Resolve part (a)

First it's important to remember the De Broglie equation and Planck's relation. De Broglie's equation is represented as \(p = h/λ\) and Planck's relation is \(E = hν\), where \(ν = c/λ\). For a photon, the momentum is \(p = h/λ\), where \(h\) is Planck's constant \(6.626 \times 10^{-34}\ Js\) and \(λ\) is the wavelength. Thus, the wavelength \(λ\) can be calculated as: \(λ = h/p\). Given that momentum \(p\) is equal to the momentum of the electron moving at speed \(10^6 m/s\), and the mass \(m_e\) of an electron is \(9.1 \times 10^{-31}\ kg\), it can be computed as \(p = m_ev = 9.1 \times 10^{-31}\ kg \times 10^6\ m/s\). The value derived can be substituted back into the wavelength equation to get the actual value.

Solve part (b)

To find the kinetic energy ratio of the photon to the electron, two formulas need to be used: the energy formula for a photon, \(E = hν\), and the kinetic energy formula for a particle, \(KE = 1/2 mv^2\). Here, \(h\) is Planck's constant, \(ν\) is frequency, \(m\) is the mass of electron, and \(v\) is the electron speed. Calculate the kinetic energy of the electron \(KE_e = 1/2 \times 9.1 \times 10^{-31}\ kg \times (10^6\ m/s)^2\). Using Planck-Einstein equation, the energy of the photon is \(E = pc\), where \(p\) is the momentum and \(c\) is the speed of light (\(3 \times 10^8\ m/s\)). Therefore, \(E = 9.1 \times 10^{-31}\ kg \times 10^6\ m/s \times 3 \times 10^8\ m/s\). Now, the ratio of kinetic energies can be calculated by dividing the kinetic energy of the photon by kinetic energy of the electron.

Key concepts

De Broglie Wavelength

In the realm of quantum mechanics, particles like electrons have wave-like properties. This is elegantly captured by the De Broglie wavelength. According to the De Broglie hypothesis, any moving particle or object has an associated wavelength. This wavelength is given by the equation:\[\lambda = \frac{h}{p}\]Where:

• \(\lambda\) is the wavelength.
• \(h\) is Planck's constant, \(6.626 \times 10^{-34}\) Js.
• \(p\) is the momentum of the particle.

The momentum \(p\) is the product of mass and velocity for objects with mass. But for photons, which have no rest mass, the momentum is instead derived from their energy. This concept bridges the gap between traditional physics, where particles are seen as point-mass objects, and modern physics, where wave and particle properties converge. When calculating the De Broglie wavelength for a photon, it's crucial to initially equate the photon's momentum with that of another particle, like an electron, as in our example exercise. This leads to meaningful comparisons and deeper insights into particle-wave symmetry.

Electron Kinetic Energy

Kinetic energy is a concept that describes the energy of an object in motion. For an electron moving at high speeds, such as in our exercise, this energy is calculated using the standard kinetic energy formula:\[KE = \frac{1}{2}mv^2\]Where:

• \(KE\) is the kinetic energy.
• \(m\) is the mass of the electron, approximately \(9.1 \times 10^{-31}\) kg.
• \(v\) is the velocity of the electron.

This equation illustrates that kinetic energy is directly proportional to both the mass of the particle and the square of its velocity. Means an increase in speed results in significantly more energy.Electrons, being subatomic, gain considerable kinetic energy even at speeds less than the speed of light, enhancing their wave nature. When comparing the kinetic energy of electrons to photons, as with the provided exercise, understanding the distinct ways energy is stored or manifested between massless photons and massive electrons becomes apparent. It demonstrates the role mass plays in classical kinetic energy calculations.

Photon Energy

Photon energy is linked to the wave nature of photons and is primarily determined by their frequency. For photons, their energy can be calculated using the relationship given by the Planck-Einstein equation:\[E = hu\]Where:

• \(E\) is the energy of the photon.
• \(h\) is Planck's constant.
• \(u\) (nu) is the frequency of the photon.

Additionally, since frequency \(u\) is related to the speed of light \(c\) and wavelength \(\lambda\) by the equation \(u = c / \lambda\), the photon energy can also be expressed as:\[E = \frac{hc}{\lambda}\]Photons behave uniquely due to their lack of mass, which allows them to exhibit both wave-like and particle-like properties simultaneously. This duality is essential in understanding phenomena across quantum mechanics, including how light interacts with matter. In the context of our exercise, when comparing the energies of photons and electrons, it highlights how photons derive their energies from sources different than kinetic motion, e.g., changes in electromagnetic waves rather than tangible velocities.