An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature

Question

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann’s constant, 1.38×10-23 J/K).

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RI SƠ 4 years 2021-08-11T23:49:18+00:00 1 Answers 29 views 0

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  1. MichaelMet
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    2021-08-11T23:51:15+00:00
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    Answer:

    The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

    The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

    Explanation:

    Thermal kinetic energy of electron or proton = KE

    ∴ KE = 3kbT/2

    given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

    so we substitute

    KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

    kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

    Now we know that

    mass of electron M’e = 9.109 ×  10⁻³¹

    mass of proton M’p = 1.6726 ×  10⁻²⁷

    We also know that

    KE = p₂ / 2m

    from the equation, p = √ (2mKE)

    { p is momentum, m is mass }

    de Broglie wavelength = β

    so β = h / p = h / √ (2mKE)

    h = Planck’s constant = 6.626 ×  10⁻³⁴

    ∴ βe =  h / √ (2m’e × KE)

    βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

    βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

    βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

    βe = 2.443422 × 10⁻⁹ m

    βp =  h / √ (2m’p ×KE)

    βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

    βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

    βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

    βp = 5.702140 × 10⁻¹¹ m

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