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A beam of electrons is accelerated from rest through a potential difference of 0.100 \(\mathrm{kV}\) and then passes through a thin slit. The diffracted beam shows its first diffraction minima at \(\pm 11.5^{\circ}\) from the original direction of the beam when viewed far from the slit. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

Short Answer

Expert verified
(a) No, relativistic formulas are not needed. (b) The slit width is approximately \( 6.18 \times 10^{-10} \text{ m} \).

Step by step solution

01

Calculate the Electron's Speed

First, let's determine the speed of the electrons after being accelerated by the potential difference of 0.100 kV. \( V = 0.100 \times 10^3 \text{ V} = 100 \text{ V} \). The kinetic energy gained by the electrons can be given as \( KE = eV = \frac{1}{2}mv^2 \), where \( e \) is the elementary charge \( (1.60 \times 10^{-19} \text{ C}) \), \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \text{ kg}) \), and \( v \) is the speed. We can solve for speed, \( v \), using \( v = \sqrt{\frac{2eV}{m}} \). Plugging in the values gives \( v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 100}{9.11 \times 10^{-31}}} \). Calculating, we find \( v \approx 5.93 \times 10^6 \, \text{m/s} \).
02

Check for Relativistic Effects

Now, we determine whether relativistic formulas are necessary by comparing the electron's speed to the speed of light \( c \approx 3 \times 10^8 \text{ m/s} \). Relativistic effects become significant if \( v \) is a significant fraction of \( c \). Here, \( \frac{v}{c} \approx \frac{5.93 \times 10^6}{3 \times 10^8} \approx 0.02 \). Since this ratio is much less than 0.1, relativistic effects are negligible. Therefore, we don't need relativity formulas.
03

Use the Diffraction Formula

The formula for the position of the first minima in a single-slit diffraction pattern is given by \( a \sin(\theta) = m\lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the minima, \( m \) is the order of minima (1 in this case), and \( \lambda \) is the wavelength of the electrons. We already have \( \theta = 11.5^{\circ} \) and \( \sin(\theta) = \sin(11.5^{\circ}) \). We'll find \( \lambda \) in the next step.
04

Calculate the Wavelength of the Electrons

The wavelength \( \lambda \) of electrons is determined using the de Broglie relation \( \lambda = \frac{h}{mv} \), where \( h \) is Planck’s constant \( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \). Plug in the values: \( \lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 5.93 \times 10^6} \). Solving gives \( \lambda \approx 1.23 \times 10^{-10} \text{ m} \).
05

Calculate the Slit Width

Using \( a \sin(\theta) = m\lambda \), solve for the slit width \( a \): \[ a = \frac{m\lambda}{\sin(\theta)} \]. With \( m = 1 \), \( \lambda = 1.23 \times 10^{-10} \) m, and \( \sin(11.5^{\circ}) \approx 0.199 \), substitute the values: \( a \approx \frac{1.23 \times 10^{-10}}{0.199} \approx 6.18 \times 10^{-10} \text{ m} \).
06

Conclusion: Answering the Questions

For (a), since the electron's speed is only about 2% of the speed of light, relativistic formulas are not needed. For (b), the slit width is approximately \( 6.18 \times 10^{-10} \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

de Broglie Wavelength
The concept of de Broglie wavelength is incredibly fascinating as it bridges the worlds of classical and quantum physics. Named after French physicist Louis de Broglie, it proposes that all matter exhibits wave-like properties. This idea is central to the field of quantum mechanics. The de Broglie wavelength (\( \lambda \)) of a particle is given by the equation:\[\lambda = \frac{h}{mv}\]Here, \( h \) is Planck’s constant (\( 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \)), \( m \) is the mass of the particle, and \( v \) is its velocity.
This concept is particularly useful in describing particles at atomic and subatomic scales, where wave-particle duality is significant. For electrons, which are much smaller and have higher velocities, this wavelength can be calculated to determine interference and diffraction patterns. The smaller the particle, the more pronounced these effects become.
Diffraction Minima
Diffraction minima occur when waves bend around an obstacle or pass through a slit and interfere with each other. This results in points of zero intensity, known as minima, where destructive interference occurs.
When an electron beam passes through a slit, it diffracts and spreads out, producing a pattern of dark and bright regions. The first diffraction minima can be found using the formula:\[a \sin(\theta) = m\lambda\]where \( a \) is the slit width, \( \theta \) is the angle of the minima, \( m \) is the order of minima, and \( \lambda \) is the wavelength of the particles. In our scenario, with the angle given as \( 11.5^{\circ} \) and the wavelength calculated using de Broglie's equation, we can determine the slit width.
This concept explains experiments like single-slit and double-slit experiments, which show light and electrons behaving as waves.
Relativistic Effects
Relativistic effects come into play when objects move at speeds comparable to the speed of light (\( 3 \times 10^8 \text{ m/s} \)). At these speeds, classical mechanics fall short, and Einstein’s theory of relativity provides a more accurate description of physics.
For electrons accelerated through a potential difference, we calculate their speed to evaluate if relativistic effects are significant. We compare the electron's speed (\( v \)) with the speed of light, determining the ratio \( \frac{v}{c} \). If this ratio is greater than 0.1, relativistic corrections might be necessary.
In this case, the calculated speed was \( 5.93 \times 10^6 \text{ m/s} \), resulting in a very small fraction of the speed of light (\( \approx 0.02 \)). Hence, relativistic effects are negligible, allowing us to use classical formulas.
Electron Speed Calculation
Calculating the speed of electrons after acceleration through a potential difference is crucial for many experiments and applications. The energy gained by each electron when accelerated through a voltage (\( V \)) can be determined using:\[ KE = eV = \frac{1}{2}mv^2 \]where \( e \) is the elementary charge, \( m \) is the electron mass, and \( v \) is the speed of the electrons.
Rearranging to find speed, the formula is:\[ v = \sqrt{\frac{2eV}{m}} \]By substituting the known values for an electron off a potential of 100 V, we found the speed to be \( 5.93 \times 10^6 \text{ m/s} \). This calculation is pivotal for applications involving electron beams, such as electron microscopy and particle acceleration experiments.

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Most popular questions from this chapter

(a) The uncertainty in the \(y\) -component of a proton's position is \(2.0 \times 10^{-12} \mathrm{m} .\) What is the minimum uncertainty in a simultaneous measurement of the \(y\) -component of the proton's velocity? (b) The uncertainty in the \(z\) -component of an electron's velocity is 0.250 \(\mathrm{m} / \mathrm{s}\) . What is the minimum uncertainty in a simultaneous measurement of the \(z\) -coordinate of the electron?

How many photons per second are emitted by a \(7.50-\mathrm{mW}\) \(\mathrm{CO}_{2}\) laser that has a wavelength of 10.6\(\mu \mathrm{m} ?\)

Electrons go through a single slit 150 \(\mathrm{nm}\) wide and strike a screen 24.0 \(\mathrm{cm}\) away. You find that at angles of \(\pm 20.0^{\circ}\) from the center of the diffraction pattern, no electrons hit the screen but electrons hit at all points closer to the center, (a) How fast were these electrons moving when they went through the slit? (b) What will be the next larger angles at which no electrons hit the screen?

Doorway Diffraction. If your wavelength were 1.0 \(\mathrm{m}\) , you would undergo considerable diffraction in moving through a doorway. (a) What must your speed be for you to have this wave length? (Assume that your mass is 60.0 \(\mathrm{kg.}\) ) (b) At the speed calculated in part (a), how many years would it take you to move 0.80 \(\mathrm{m}\) (one step)? Will you notice diffraction effects as you walk through doorways?

(a) A nonrelativistic free particle with mass \(m\) has kinetic energy \(K .\) Derive an expression for the de Broglie wavelength of the particle in terms of \(m\) and \(K .\) (b) What is the de Broglie wavelength of an 800 -eV electron?

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