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Chapter 28: Problem 62

A double-slit interference experiment is performed with 2.0-ev photons. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (i.e., the spacing between maxima is the same)?

Short Answer

Expert verified
Answer: The kinetic energy of the electrons is approximately 2.43 × 10^{-25} Joules.

Step by step solution

01

Recall the energy-wavelength relation for photons

The energy of a photon is related to its wavelength by the equation: E = h * c / λ where E is the energy, h is the Planck's constant (6.626 × 10^{-34} J·s), c is the speed of light (3 × 10^8 m/s), and λ is the wavelength.
02

Determine the wavelength of the given photons

We are given the energy of the photons: 2.0 eV. First, we need to convert the energy from electron volts to Joules: E = 2 * 1.602 × 10^{-19} J/eV = 3.204 × 10^{-19} J Now, use the energy-wavelength relation for photons to find the wavelength: λ = h * c / E λ = (6.626 × 10^{-34} J·s) * (3 × 10^8 m/s) / (3.204 × 10^{-19} J) λ ≈ 6.21 × 10^{-10} m
03

Recall the de Broglie relation

The de Broglie relation is used for particles, such as electrons, and relates their wavelength to their momentum: λ = h / p where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.
04

Relate momentum to kinetic energy

We know that the momentum of a particle with mass m and velocity v is given by: p = m * v We also know that kinetic energy (KE) is given by: KE = (1/2) * m * v^2 We want to rewrite the momentum equation in terms of kinetic energy. First, solve the KE equation for v: v = sqrt(2 * KE/m) Now substitute this expression for v in the momentum equation: p = m * sqrt(2 * KE/m)
05

Find the kinetic energy of the electrons

We have the same wavelength for the electrons as for the photons, λ = 6.21 × 10^{-10} m. Using the de Broglie relation and the expression for momentum in terms of kinetic energy, we get: λ = h / (m * sqrt(2 * KE/m)) Now, the mass of an electron is: m_e = 9.11 × 10^{-31} kg. We can rewrite the equation and solve for KE: KE = h^2 / (2 * m_e * λ^2) Plug in the values of h, m_e, and λ to find the kinetic energy of the electrons: KE = (6.626 × 10^{-34} J·s)^2 / (2 * (9.11 × 10^{-31} kg) * (6.21 × 10^{-10} m)^2) KE ≈ 2.43 × 10^{-25} J So, the kinetic energy of the electrons is about 2.43 × 10^{-25} Joules.

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Most popular questions from this chapter

The particle in a box model is often used to make rough estimates of energy level spacings. Suppose that you have a proton confined to a one-dimensional box of length equal to a nuclear diameter (about \(10^{-14} \mathrm{m}\) ). (a) What is the energy difference between the first excited state and the ground state of this proton in the box? (b) If this energy is emitted as a photon as the excited proton falls back to the ground state, what is the wavelength and frequency of the electromagnetic wave emitted? In what part of the spectrum does it lie? (c) Sketch the wave function \(\psi\) as a function of position for the proton in this box for the ground state and each of the first three excited states.
What is the maximum possible value of the angular momentum for an outer electron in the ground state of a bromine atom?
A hydrogen atom has a radius of about \(0.05 \mathrm{nm}\) (a) Estimate the uncertainty in any component of the momentum of an electron confined to a region of this size. (b) From your answer to (a), estimate the electron's kinetic energy. (c) Does the estimate have the correct order of magnitude? (The ground-state kinetic energy predicted by the Bohr model is \(13.6 \mathrm{eV} .\) )
The energy-time uncertainty principle allows for the creation of virtual particles, that appear from a vacuum for a very brief period of time $\Delta t,\( then disappear again. This can happen as long as \)\Delta E \Delta t=\hbar / 2,\( where \)\Delta E$ is the rest energy of the particle. (a) How long could an electron created from the vacuum exist according to the uncertainty principle? (b) How long could a shot put with a mass of \(7 \mathrm{kg}\) created from the vacuum exist according to the uncertainty principle?
An x-ray diffraction experiment using 16 -keV x-rays is repeated using electrons instead of \(x\) -rays. What should the kinetic energy of the electrons be in order to produce the same diffraction pattern as the x-rays (using the same crystal)?
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