Question

Coherent light is passed through two narrow slits whose separation is 20.0 $\mu$m. The second-order bright fringe in the interference pattern is located at an angle of 0.0300 rad. If electrons are used instead of light, what must the kinetic energy (in electron volts) of the electrons be if they are to produce an interference pattern for which the second-order maximum is also at 0.0300 rad?

Short answer

The electrons must have a kinetic energy of approximately 9,432 eV.

Step-by-step solution

Understanding the Double Slit Experiment

In a double-slit experiment, coherent waves such as light or electron waves create interference patterns through constructive interference, leading to bright fringes. The positions of these fringes depend on the wavelength of the incident light or particles and the separation between slits.

Using the Interference Formula for Bright Fringes

The formula for the angle for bright (constructive interference) fringes is given by: $$d\sin (\theta )=m\lambda$$where $d$ is the distance between the slits, $\theta$ is the angle with the central maximum, $m$ is the order of the maximum (integer), and $\lambda$ is the wavelength.

Calculating the Wavelength for the Second-Order Maximum

Given $m=2$ (second order), $\theta =0.0300$ rad, and $d=20.0\mu m=20.0\times {10}^{-6}$ m:Substitute into the interference formula:$$20.0\times {10}^{-6}\sin (0.0300)=2\lambda$$Solving for $\lambda$, we find $\lambda =3.0\times {10}^{-8}$ m.

Relating Electron Wavelength to Kinetic Energy

The de Broglie wavelength formula relates the wavelength $\lambda$ of a particle to its momentum $p$:$$\lambda =\frac{h}{p}$$where $h=6.626\times {10}^{-34}$ Js is Planck's constant. For electrons, momentum can be expressed as $p=\sqrt{2m_{e}K}$, with $m_e=9.109\times {10}^{-31}$ kg (electron mass) and $K$ as kinetic energy.

Solving for Electron Kinetic Energy

Substitute for momentum in the de Broglie formula:$$\lambda =\frac{h}{\sqrt{2m_{e}K}}$$Rearrange to solve for $K$:$$K=\frac{h^2}{2m_e{\lambda }^2}$$Substitute $\lambda =3.0\times {10}^{-8}$ m and known constants:$$K=\frac{(6.626\times {10}^{-34}{)}^2}{2\times 9.109\times {10}^{-31}\times (3.0\times {10}^{-8}{)}^2}$$Calculating gives $K=1.51\times {10}^{-3}$ Joules, which converts to electron volts (1 eV = 1.602 \times 10^{-19} J):$$K=\frac{1.51\times {10}^{-3}}{1.602\times {10}^{-19}}\approx 9,432\text{ eV}$$

Final Answer

The kinetic energy of the electrons needed to produce the same second-order maximum at an angle of 0.0300 rad is approximately $9,432$ eV.

Key concepts

Interference Patterns

The concept of interference patterns is central to understanding the double slit experiment. These patterns occur when waves overlap, creating regions of constructive and destructive interference. Constructive interference happens when the peaks of waves coincide, resulting in amplified brightness, while destructive interference occurs when peaks and troughs collide, canceling each other out. This phenomenon was historically observed with light, leading to a series of bright and dark regions called fringes.
Imagine a pond where two stones are dropped at once. The overlapping ripples create a similar pattern to that of the interference pattern seen through the slits. This is a hallmark of wave behavior and is evidence of wave-like properties in particles such as electrons.

de Broglie Wavelength

The de Broglie wavelength bridges the wave-particle duality of matter, proposing that particles have wave properties. This is given by the formula: $$\lambda =\frac{h}{p}$$ where $\lambda$ is the wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle. This concept was groundbreaking in physics because it introduced the idea that not just light, traditionally viewed as a wave, but also particles such as electrons have wavelengths. Understanding an electron's de Broglie wavelength helps explain phenomena like diffraction and interference when they pass through the slits.

Electron Kinetic Energy

To determine the kinetic energy of electrons in experiments like those involving interference patterns, we use the relationship between kinetic energy and momentum given by the de Broglie and classical physics, which is integrated in the formula: $$K=\frac{h^2}{2m_e{\lambda }^2}$$ where $K$ is the kinetic energy, $m_e$ is the mass of the electron, and $\lambda$ is the de Broglie wavelength. Calculating this gives the energy in joules but we often express it in electron volts (eV) for convenience in quantum experiments. In this scenario, knowing the wavelength helps us find the specific energy electrons need to create a given interference pattern, providing a link between theory and experimental evidence.

Constructive Interference

Constructive interference occurs when the crests of two waves align closely, making the resultant wave amplitude larger. In the context of the double slit experiment, this is seen as bright fringes on the interference pattern.
The formula for the condition of constructive interference is: $$d\sin (\theta )=m\lambda$$ where $d$ is the distance between slits, $\theta$ is the angle of observation, $m$ is the order of the maximum (whole number), and $\lambda$ is the wavelength. Recognizing the conditions for constructive interference helps predict where the bright spots form and aligns with understanding the underlying wave mechanics of particles. This ability to predict and confirm the existence of interference patterns through constructive interference is fundamentally significant in verifying wave-particle duality.