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Chapter 28: Problem 67

A beam of electrons is accelerated across a potential of \(15 \mathrm{kV}\) before passing through two slits. The electrons form a interference pattern on a screen \(2.5 \mathrm{m}\) in front of the slits. The first-order maximum is \(8.3 \mathrm{mm}\) from the central maximum. What is the distance between the slits?

Short Answer

Expert verified
#Answer# After finding the electron's velocity, calculating the de Broglie wavelength, determining the angle of electron's interference, and using the formula for electron interference, we can determine the distance between the two slits (d) in the problem.

Step by step solution

01

Find the electron's velocity after being accelerated

First, we need to find the velocity of the electrons after being accelerated through the given potential difference of \(15 \, \mathrm{kV}\). The kinetic energy gained by each electron is equal to the potential energy it loses. Therefore, we have: \(eV = \frac{1}{2}mv^2\) Where \(e\) is the elementary charge (\(1.6 \times 10^{-19} \, \mathrm{C}\)), \(V\) is the potential difference (\(15 \, \mathrm{kV}\)), \(m\) is the mass of an electron (\(9.11 \times 10^{-31} \, \mathrm{kg}\)), and \(v\) is the electron's velocity. We can solve for the electron's velocity: \(v = \sqrt{\frac{2eV}{m}}\)
02

Calculate the de Broglie wavelength of the electrons

Now that we have the electron's velocity, we can find its de Broglie wavelength using the formula: \(\lambda = \frac{h}{mv}\) Where \(\lambda\) is the de Broglie wavelength, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \, \mathrm{Js}\)), \(m\) is the mass of the electron, and \(v\) is the electron's velocity.
03

Determine the angle of electron's interference

To find the angle of electron's interference for the first-order maximum, we will use the following formula: \(\sin(\theta) = \frac{L_y}{L_x}\) Where \(\theta\) is the angle of interference, \(L_y\) is the distance of the first-order maximum from the central maximum (\(8.3 \, \mathrm{mm}\)), and \(L_x\) is the distance between the screen and the slits (\(2.5 \, \mathrm{m}\)). First, we will convert the distance of the first-order maximum to meters: \(8.3\, \mathrm{mm} = 0.0083 \, \mathrm{m}\). Then we will calculate the angle: \(\theta = \arcsin \left( \frac{L_y}{L_x} \right)\)
04

Calculate the distance between the two slits

Now that we have the de Broglie wavelength of the electrons and the angle of interference, we can use the formula for electron interference to find the distance between the two slits: \(d \sin(\theta) = m \lambda\) Where, \(d\) is the distance between the slits, \(\theta\) is the angle of interference, \(m\) is the order of the maximum (1 for first-order), and \(\lambda\) is the de Broglie wavelength. We can solve for the distance between the slits: \(d = \frac{m \lambda}{\sin(\theta)}\) Now, substituting the values, and solving for \(d\), we will get the distance between the slits which is the required solution.

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The omega particle \((\Omega)\) decays on average about \(0.1 \mathrm{ns}\) after it is created. Its rest energy is 1672 MeV. Estimate the fractional uncertainty in the \(\Omega\) 's rest energy \(\left(\Delta E_{0} / E_{0}\right)\) [Hint: Use the energy-time uncertainty principle, Eq. \((28-3) .]\)
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