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pfavPHYS 212-071 HW-3 Solutionzhbq

By Dean Hunt,2014-10-25 08:27
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pfavPHYS 212-071 HW-3 Solutionzhbq

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    PHYS 212-071

    Solutions to HW-3

    nd Edition) (Numbers refer to 2

    Pb.16 Consider the metals lithium, beryllium, and mercury, which have work functions of 2.3 eV, 3.9 eV, and 4.5 eV, respectively. If light of wavelength 300 nm is incident on each of these metals, determine (a) which metal exhibits the photoelectric effect and (b) the maximum kinetic energy for the photoelectron in each case.

    A. 16 The metals which exhibit the photoelectric effect are those for which the maximum kinetic energy of the ejected electrons is positive. Namely

     Kh0max

    8c3~10m/s1515We have eV.s and. Therefore we get the 10Hzh4.136~109(300~10m

    following table:

    Metal [eV] K[eV] Exhibit Photoelectric Effect max

    Lithium 2.3 1.836 YES

    Beryllium 3.9 0.236 YES

     0.364Mercury 4.5 NO

    Pb. 20. Figure P2.20 shows the stopping potential versus incident photon frequency for the photoelectric effect for sodium. Use these data points to find (a) the work function, (b) the ratio, and (c) the cutoff wavelength. (d) Find the percent difference between your h/e

    answer to (b) and the accepted value.

    A 20 (a) The work function should be found using the graph by extrapolating the linear curve back to zero frequency. At this point we have. Knowing the electronic V/es

    charge, we can deduce the work function eV[eV]. (b) Similarly the ratio is to h/es

    Vhsbe found from the slope of the linear graph, namely [V.s]. Knowing the ef

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    . (c) The cutoff electronic charge we can find Planck’s constant from the graph hgraph

    wavelength is the wavelength for which

    hchc1240.8[eV.nm]. Khf0.?[nm](max00[eV](0

    |hh|hgraphaccepted(d) The % difference can be found by calculating [%]100~hhaccepted

    Pb. 25. X-rays with a wavelength of 0.040 nm undergo Compton scattering. (a) Find the

    oooooowavelength of photons scattered at angles of 30, 60, 120, 150, 180, and 210. (b) Find

    the energy of the scattered electrons corresponding to these scattered x-rays. (c) Which one the given scattering angles provides the electron with the greatest energy?

    A 25. (a) The wavelengthsof the scattered photons can be calculated using the (

    hhCompton scattering formula, where nm is the ;;((1cos0.002430mcmcee

    Compton wavelength of the electron. The kinetic energies of the scattered electrons can

    hchcbe found using the following formulaKhh. We therefore get the e0((0

    following table:

     [degrees] 30 60 120 150 180 210

    [nm] (0.000326 0.001215 0.003645 0.004534 0.00486 0.004534

     [nm] (0.040326 0.041215 0.043645 0.044534 0.04486 0.044534

    [eV] Ke250 914 2591 3158 3361 3158

    oThe scattering angle of 180 provides the electron with the greatest kinetic energy of 3.4 keV.

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    Pb. 32. Show that Compton formula

    h ;;((1cosmce

    results when expressions for the electron energy (Equation 2.34) and momentum (Equation 2.35) are substituted into the relativistic energy expression,

    22224 Epcmceee

    2A 32. Equation 2.34 gives the (total) electron energy:. Equation Ehfhfmcee

    222hfhfhff2??2p(2.35) gives the electron’s momentum:. By cos????e2ccc??

    substitution of these expressions into the relativistic energy expression we get:

    222(2hfhf2hff??2224 ;;hfhfmccos~cmc????)?ee2ccc??)???

    Expanding the square on the left-hand side of the equation and after eliminating equal terms we get:

    h ;;ffff1cos2mce

    ccc(Multiplying through by, and recalling that (and, we get ffff

    h ;;((1cosmce

    QED.

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