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23 Nov 2019

An electron and a proton each have a thermal kinetic energy of3kBT/2. Calculate the de Broglie wavelength of each particle at atemperature of 2040 K. (kB is Boltzmann's constant, 1.38×10-23 J/K.The mass of the electron is 9.11×10-31 kg and the mass of theproton is 1.67×10-27 kg.)
Wavelength of the electron =
Wavelength of the proton =

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Jean Keeling
Jean KeelingLv2
6 Jul 2019
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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

The neutron has a mass of 1.67 10-27 kg. Neutrons emitted in nuclear reactions can be slowed down by collisions with the matter. They are referred to as thermal neutrons after they come into thermal equilibrium with the environment. The average kinetic energy ( 3/2 kBT) of a thermal neutron is approximately 0.04 eV. Calculate the de Broglie wavelength of a neutron with a kinetic energy of 0.04 eV?

A particle moves about in a one-dimensional box of length L. Its energy is quantized by the condition n(?/2) = L, where ? is the de Broglie wavelength of the particle. (a) Show that the allowed energies are given by En = n2E1, where E1 = h28mL2 and m is the mass of the particle. Evaluate E1 for (b) an electron in a box of size L = 0.1 nm and (c) a proton in a box of size L = 10^-15 m. (Take the potential energy of the particle in the box to be zero so that its total energy is its kinetic energy p2/2m.)

In 1924, Louis de Broglie postulated that particles such aselectrons and protons might exhibit wavelike properties. Histhinking was guided by the notion that light has both wave andparticle characteristics, so he postulated that particles such aselectrons and protons would obey the same wavelength-momentumrelation as that obeyed by light: \lambda = h/p, where lambda isthe wavelength, p the momentum, and h Planck's constant.

a. Find the de Broglie wavelength lambda for an electron moving ata speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed islow enough that the classical momentum formula p=mv is stillvalid.) Recall that the mass of an electron is m_{\rm e} =9.11\times 10^{-31}\; {\rm kg}, and Planck's constant is h = 6.626\times 10^{-34}\; {\rm J \cdot s}.
Express your answer in meters to three significant figures.
lambda =7.270×10-10 \rm m

b. Find the de Broglie wavelength lambda of a baseball pitched at aspeed of 43.7 m/s. Assume that the mass of the baseball is 0.143\;{\rm kg}.
Express your answer in meters to three significant figures
lambda =1.06×10-34 \rm m

c. Consider a beam of electrons in a vacuum, passing through a verynarrow slit of width 2.00 \;\mu{\rm m}. The electrons then headtoward an array of detectors a distance 1.076 m away. Thesedetectors indicate a diffraction pattern, with a broad maximum ofelectron intensity (i.e., the number of electrons received in acertain area over a certain period of time) with minima of electronintensity on either side, spaced 0.518 cm from the center of thepattern. What is the wavelength lambda of one of the electrons inthis beam? Recall that the location of the first intensity minimain a single slit diffraction pattern for light is y=L \lambda /a,where L is the distance to the screen (detector) and a is the widthof the slit. The derivation of this formula was based entirely uponthe wave nature of light, so by de Broglie's hypothesis it willalso apply to the case of electron waves.
Express your answer in meters to three significant figures.

I only need C... please answer fully, thank you

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