An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the

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An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron’s probability of tunneling through the barrier increase

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Tài Đức 3 years 2021-08-24T02:19:25+00:00 1 Answers 3 views 0

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  1. binhan
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    2021-08-24T02:20:44+00:00
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    Answer:

    factor that the electron’s probability of tunneling through the barrier increase 2.02029

    Explanation:

    given data

    kinetic energy = 10.1 eV

    height = 18.2 eV

    width = 1.00 nm

    wavelength = 546 nm

    solution

    we know that probability of tunneling is express as

    probability of tunneling = e^{-2CL}   ……………..1

    here C is = \frac{\sqrt{2m(U-E}}{h}

    here h is Planck’s constant

    c = \frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}  

    c = 2319130863.06

    and proton have hf = \frac{hc}{\lambda } = {1240}{546} = 2.27 ev

    so electron K.E = 10.1 + 2.27

    KE = 12.37 eV

    so decay coefficient inside barrier is

    c’ = \frac{\sqrt{2m(U-E}}{h}

    c’ = \frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}  

    c’ = 1967510340

    so

    the factor of incerease in transmisson probability is

    probability = e^{2L(c-c')}

    probability = e^{2\times 1\times 10^{-9} \times (351620523.06)}

    factor probability = 2.02029

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