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Find the de-Broglie wavelength of an ele...

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Text Solution

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`lambda=(h)/(sqrt(2mE))`
`=(6.625 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(31) xx 120 xx 1.6 xx 10^(-19)))`
`=1.121 xx 10^(-10)`m.
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Knowledge Check

  • The de Broglie wavelength of an electron in the first Bohr orbit is

    A
    equal to halff the circumference of the first orbit
    B
    equal toone fourth the circumference of the first orbit
    C
    equal to the circumference of the first orbit
    D
    equal to twice the circumference of the first orbit.

  • de-Broglie wavelength associated with an electron having kinetic energy 500 eV is :

    A
    `0.55Å`
    B
    `0.69Å`
    C
    `0.78Å`
    D
    `1.31Å`

  • If an electron and a proton have the same de - Broglie wavelength , then the kinetic energy of the electron is

    A
    more than that of a proton
    B
    equal to that of a proton
    C
    zero
    D
    less than that of a proton

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