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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

An electron is confined to a one-dimensional box of length \(L\). When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 0.20 eV. (a) What is the ground-state energy (in electron-volts) of the electron in this box? (b) What are the energies (in electron-volts) of the photons that can be emitted when the electron starts in its third excited state and makes transitions downwards to the ground state (either directly or through intervening states)? (c) Sketch the wave function of the electron in the third excited state. (d) If the box were somehow made longer, how would the electron's new energy level spacings compare with its old ones? (Would they be greater, smaller, or the same? Or is more information needed to answer this question? Explain.)
To resolve details of an object, you must use a wavelength that is about the same size, or smaller, than the details you want to observe. Suppose you want to study a molecule that is about \(1.000 \times 10^{-10} \mathrm{m}\) in length. (a) What minimum photon energy is required to study this molecule? (b) What is the minimum kinetic energy of electrons that could study this? (c) Through what potential difference should the electrons be accelerated to reach this energy?
List the number of electron states in each of the subshells in the \(n=7\) shell. What is the total number of electron states in this shell?
The distance between atoms in a crystal of \(\mathrm{NaCl}\) is $0.28 \mathrm{nm} .$ The crystal is being studied in a neutron diffraction experiment. At what speed must the neutrons be moving so that their de Broglie wavelength is \(0.28 \mathrm{nm} ?\)
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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