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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 circyumference of the first orbit

C

Equal to twice the circumference of the first orbit

D

Equal to the circulference of the first orbit

Text Solution

Verified by Experts

The correct Answer is:
D
`mvr_(n) = (nh)/(2pi) rArr pr_(n) = (nh)/(2pi) rArr (h)/(lambda) xx r_(n) = (nh)/(2pi)`
`rArr lambda = (2pir_(n))/(n)`, for first orbit `n = 1` so `lambda = 2pi r_(1)`
`=` circumference of first 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`

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    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 `

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