Calculate the de Broglie wavelength for an electron in the first Bohr orbit of a hydrogen atom?

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  • 9 years ago
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    According to deBroglie, λ = the circumference (2πr) of orbit for n=1

    Finding this circumference ..

    Total energy (E) of an electron in a hydrogen atom .. E = (-) ke² / 2r

    For the ground state (n=1) this energy is -13.60 eV

    E = (13.60 x 1.60^-19)J .. .. E = 2.176^-18 J

    2.176^-18 = ke² / 2r

    r = ke² / (2 x 2.176^-18)

    r = (9.0^9)(1.60^-19)² / (4.352^-18) .. .. r = 5.29^-11 m

    According to deBroglie 1λ = the circumference of orbit ..

    So .. λ = 2πr = 2 x π x 5.29^-11m .. .. .. ► λ = 3.326^-10 m

  • 4 years ago

    Classical (Newtonian) physics in maximum circumstances is barely an approximation. that does no longer mean it somewhat isn't any longer sensible. The Bohr style makes ultimate prediction - on the subject of the dimensions of the hydrogen atom, valence stages, and how lots power is mandatory to get rid of an electron. moving from Bohr's style to sparkling up misconceptions is larger than beginning with a quantum style young ones won't be waiting to fully carry close. id, on the different hand, posits an all useful "fashion designer", which predicts something, and subsequently explains no longer something. in actuality, any declare that could't be falsified explains no longer something. id isn't technological information. besides, in accordance to quantum physics radioactive decay is an uncaused technique. which potential it somewhat is a threat no longer each thing has a reason, and the reasoning used to declare that the universe or life inevitably might desire to be brought about is going out the window.

  • 9 years ago

    From Standard formulation it is clear that the velocity of electron in Bhor's first orbit = 2.18*10^6 m/s

    Now de Broglie wavelength is given by lamda = h/mv

    h = Plank's Constant = 6.626*10^-34 Js

    m= Mass of an electron = 9.1*10^-31 Kg

    v = Velocity of an electron.

    Using this lamda = 3.3366*10^-10 meter = 3.3366 Armstrong.

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