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The de-Broglie wavelength of an electron...

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

A

equal to one- fourth the circumference of the first orbit

B

equal to half the circumference of first orbit

C

equal to twice the circumference of first orbit.

D

equal to the circumference of the first orbit.

Text Solution

Verified by Experts

The correct Answer is:
D
Angular momentum `=(nh)/(2pi)`
`rArr` moment of momentum `=pxxr_n`
`rArrpxxr_n=(nh)/(2pi)rArrr(h)/(lambda)r_nrArrlambda =(2pir_n)/(n)`
For `1^st` orbit, `n=1,lambda =2pir_1`
`rArr =` circumference of `1^st` orbit.
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Knowledge Check

  • The de Broglie wavelength of an electron in the nth Bohr orbit is related to the radius R of the orbit as:

    A
    `nlambda=nR`
    B
    `nlambda=3/2piR`
    C
    `nlambda=2piR`
    D
    `nlambda=4piR`

  • The de Broglie wavelength of an electron in the 3rd Bohr orbit is

    A
    `2pia_0`
    B
    `4pia_0`
    C
    `6pia_0`
    D
    `8pia_0`

  • The de Broglie wavelength of an electron in the n^(th) Bohr orbit is related to the radius R of the orbit as

    A
    `n lambda = pi R `
    B
    `n lambda = (3)/(2) pi R `
    C
    ` n lambda = 2 pi R `
    D
    ` n lambda = 4 pi R `