PHYS 1070 Lecture Notes - Lecture 7: European Route E6, Electron Configuration, Matter Wave

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9 Aug 2018

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Wave part: = c / f. An em wave is produced when an electrical charge is accelerated. Em waves (light) is released in individual lumps or quanta of energy we revert as photons. E = hf = h(c / ) Standing wave condition: l = (n ) / 2. = c / f = hc / e. = h / p = h / mv. Every bond consists of two electrons. There are regions where they are more likely to be. N number of half wavelengths, this is the electronic state de broglie wavelength. Ke = e = mv = p /2m. Charge: e = 1. 6 10 c. Energy depends on quantum number n (= 1, 2, 3 ) Spin of the electron also has a quantum number s. In a given system, no two electrons can occupy the same state. E s will occupy the lowest energy levels possible first (two per -bond)

Related Questions

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 =7.270Ã10-10 \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 =1.06Ã10-34 \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

PHYS 1070 Lecture Notes - Lecture 7: European Route E6, Electron Configuration, Matter Wave

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