Question

An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy \(K\) of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron?

Short answer

(a) \( E_p = \frac{hc}{\lambda_e} = K \frac{\lambda_e}{c} \). (b) The photon's energy \( E_p \) is greater.

Step-by-step solution

Understanding the Problem

We have electron and photon beams passing through identical slits, and the first dark fringe occurs at the same angle for both beams. We need to find a relationship between the energy of a photon and the kinetic energy of an electron, and determine which energy is greater.

Use Diffraction Condition

For a slit, the dark fringe condition is given by \( d \sin \theta = m\lambda \) (for the first order, \( m = 1 \)). Since the angle \( \theta \) is the same for both beams, their wavelengths are related by \( \lambda_e = \lambda_p \) where \( \lambda_e \) is the electron's de Broglie wavelength and \( \lambda_p \) is the photon's wavelength.

Relate the Wavelengths to Energy

For a photon, the energy \( E_p \) is given by \( E_p = \frac{hc}{\lambda_p} \). For an electron, the de Broglie wavelength is \( \lambda_e = \frac{h}{p} \), where \( p = \sqrt{2mK} \) and the kinetic energy \( K \) can be expressed as \( K = \frac{h^2}{2m\lambda_e^2} \). Because \( \lambda_e = \lambda_p \), we have \( \lambda_e = \lambda_p \).

Compare Photon and Electron Energies

Since \( \lambda_e = \lambda_p \), the energy of a photon \( E_p = \frac{hc}{\lambda_e} \) can be compared to the kinetic energy of an electron \( K = \frac{h^2}{2m\lambda_e^2} \). Multiply \( K \) by \( \lambda_e \) and compare to \( E_p \). Compute \( E_p = \frac{hc}{\lambda_e} = \frac{h^2c}{h}\frac{c}{K} \) implying \( E_p > K \).

Deduce Energy Relationships

Calculate ratios finallly, we express the photon's energy \( E_p \) and the electron's kinetic energy \( K \). By comparing expressions, \( E_p = \frac{hc}{\lambda} \) and \( K = \frac{h^2}{2m\lambda^2} \). Since \( hc \) is far greater than \( \frac{h^2}{2m} \), it is clear that \( E_p > K \).

Key concepts

Diffraction

Diffraction occurs when waves bend around the edges of an obstacle or spread after passing through a slit. This phenomenon is essential in understanding the behavior of both light (photons) and electrons as they encounter barriers. When a wave passes through a narrow opening, it bends and creates patterns of constructive and destructive interference on the other side. These patterns include bright and dark fringes, which can be observed on a screen placed at a distance.

In the scenario involving electron and photon beams, when both pass through identical slits, the diffraction pattern reveals crucial information about their wave nature. The condition for the first dark fringe in a diffraction pattern is given by the formula:

• For a slit, \[ d \sin \theta = m\lambda \], where \( d \) is the slit width and \( m \) is the order of the fringe.
• Here, \( \theta \) is the angle at which the dark fringe appears.

Since the angle is the same for both beams, the wavelengths related to electrons and photons can be inferred to be equivalent in this context. This link lets us explore deeper properties like energy.

Photon Energy

Photon energy is an attribute of light's particle-like characteristics. It is quantified using the equation:\[ E_p = \frac{hc}{\lambda} \]where \( E_p \) represents the photon's energy, \( h \) is the Planck constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.

This expression shows a direct relationship between the energy of a photon and its wavelength; shorter wavelengths carry more energy. Knowledge of photon energy helps us understand not only fundamental physics but also practical applications like solar energy conversion.

In our exercise, we learn that the photon's wave properties exhibit energy that is significantly larger than that of an electron's kinetic energy. The equation indicates that since the speed of light and Planck's constant remain major constants, the energy connected with light surpasses the measured electron energies when dealing with similar wavelength scenarios.

Electron Kinetic Energy

Electron kinetic energy relates to the motion and particle-like aspects of electrons. It is fundamentally tied to an electron's velocity and mass. This energy can be extracted by the relationship:\[ K = \frac{h^2}{2m\lambda^2} \]where \( K \) is the kinetic energy, \( h \) is the Planck constant, \( m \) is the electron's mass, and \( \lambda \) represents the electron's de Broglie wavelength.

Here, the energy depends on the electron's mass and wavelength. The kinetic energy is generally smaller due to the mass of an electron being quite low compared to light's velocity. When diffraction is observed, the wave-like nature of electrons is prominent, leading to calculations that link their momentum with wavelength.

Comparatively, this exercise shows that even though electrons share identical diffraction conditions—and thus wavelengths—with photons, their kinetic energy is lesser than the photon energy. This outcome aligns with the interpretation of wave-particle duality, where different particles have distinct energy interactions within similar wave conditions.