Question

A beam of electrons with a speed of \(5.25 \times 10^{4} \mathrm{m} / \mathrm{s}\) is incident on a slit of width \(200 .\) nm. The distance to the detector plane is chosen such that the distance between the central maximum of the diffraction pattern and the first diffraction minimum is \(0.300 \mathrm{cm} .\) How far is the detector plane from the slit?

Short answer

The distance between the slit and the detector plane is approximately \(0.024\:m\) or \(2.4\:cm\).

Step-by-step solution

Find the de Broglie wavelength of the electrons

Using the de Broglie wavelength formula, we can find the wavelength of the electrons: \[ \lambda = \frac{h}{mv} \] where \(h\) is Planck's constant \((6.626 × 10^{-34}\:J\:s)\), \(m\) is the mass of an electron \((9.109 × 10^{-31}\:kg)\) and \(v\) is the given speed of the electrons \((5.25 \times 10^4\:m/s)\). Now we can calculate the wavelength: \[ \lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 5.25 \times 10^4} \]

Calculate the wavelength of the electrons

After completing the calculation, we find the wavelength of the electrons: \[ \lambda \approx 1.351 \times 10^{-10}\:m \]

Apply the single-slit diffraction formula

The formula for single-slit diffraction is given by: \[ \sin(\theta) = \frac{m\lambda}{a} \] where \(m\) is the order of the diffraction patterns, \(\lambda\) is the wavelength of the electrons, \(a\) is the width of the slit, and \(\theta\) is the angle between the central maximum and the diffraction minimum. In this problem, the distance between the central maximum and the first diffraction minimum is given, i.e. the first order (\(m = 1\)). We are also given the value of \(a\), i.e. \[ a = 200\:nm = 2 \times 10^{-7}\:m \] Hence, using the given values, we can calculate the angle \(\theta\): \[ \sin(\theta) = \frac{1.351 \times 10^{-10}}{2 \times 10^{-7}} \] \[ \theta \approx 0.712^{\circ} \]

Applying trigonometry

The distance (\(d\)) between the slit and the detector plane can be found by applying simple trigonometry: \[ \tan(\theta) = \frac{b}{d} \] where \(b = 0.3\:cm = 3 \times 10^{-3}\:m\), and \(d\) is the distance we need to find. Rearranging the formula: \[ d = \frac{b}{\tan(\theta)} \] Now we plug in the values we have for \(b\) and \(\theta\): \[ d = \frac{3 \times 10^{-3}}{\tan(0.712^{\circ})} \]

Calculate the distance between the slit and detector plane

After completing the calculation, we find the distance between the slit and the detector plane: \[ d \approx 0.024\:m \] Hence, the distance between the slit and the detector plane is approximately \(0.024\:m\) or \(2.4\:cm\).

Key concepts

Understanding de Broglie Wavelength

The concept of the de Broglie wavelength unites the worlds of quantum mechanics and wave-particle duality. It postulates that every moving particle possesses a wavelength, a theory proposed by Louis de Broglie in 1924. The de Broglie wavelength is calculated using the formula:

• \( \lambda = \frac{h}{mv} \)

Here, \(h\) represents Planck's constant, \(m\) is the mass of the particle (in this case, an electron), and \(v\) is its velocity. This equation emphasizes how particle momentum (mass times velocity) inversely affects the wavelength of that particle, leading to a smaller wavelength at higher momentum.

For electrons, due to their small mass and the immense influence of quantum effects, their de Broglie wavelength is significantly noticeable. In practical terms, understanding an electron's wavelength allows us to explore phenomena like diffraction and interference, making it crucial in fields like electron microscopy.

Unveiling Single-Slit Diffraction with Electrons

Single-slit diffraction is a fascinating phenomenon where waves spread out after passing through a narrow opening, highlighting their wave-like characteristics. This diffraction occurs when the width of the slit (denoted as \(a\)) is comparable to the wavelength of the wave.For electrons, whose behavior as waves is predicted by their de Broglie wavelength, passing through a narrow slit leads to the creation of a diffraction pattern. The intensity and positioning of these waves can be calculated using the formula:

• \( \sin(\theta) = \frac{m\lambda}{a} \)

Here, \(\theta\) is the angle relative to the central maximum, \(m\) represents the order of the diffraction pattern, and \(\lambda\) is the electron's wavelength. This relationship allows for calculating specific details about the spread of the electron beam after it passes through the slit.

In educational settings, understanding single-slit diffraction helps to visualize how electrons display both particle-like and wave-like properties. It underscores the essence of quantum mechanics, illustrating that even particles traditionally thought to be solid can exhibit behaviors typical of waves.

The Nature of Electron Beams

Electron beams are streams of electrons traveling at set velocities, finding extensive use in technology and industry. In devices such as cathode ray tubes and electron microscopes, these beams are vital. Unlike light, whose photons possess negligible mass, electrons have tangible mass and charge, making their behavior more complex. These beams, when manipulated, can illustrate quantum phenomena such as interference and diffraction. For example, when an electron beam encounters obstacles, it can show diffraction patterns, mimicking the tendencies of waves despite being made up of electrons. This wave-particle duality is a cornerstone of quantum mechanics.

To put it simply, electron beams facilitate the exploration of quantum effects, providing deeper insights into the underlying principles of the micro-world. They are a remarkable tool for understanding and visualizing the dual nature of particles, as specified by quantum theory.