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The de-Broglie wavelength (lambda(B)) as...

The de-Broglie wavelength `(lambda_(B))` associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state `(lambda_(G))` by :

A

`lambda_B=3lambda_G`

B

`lambda_B=2lambda_G`

C

`lambda_B=3lambda_(G//3)`

D

`lambda_B=3lambda_(G//2)`

Text Solution

Verified by Experts

The correct Answer is:
A
If P is the linear momentum then de Broglie wavelength is `lambda=h/P`
`rArr lambda_B/lambda_G=P_G/P_B=(mv_G)/(mv_B)`
`v prop z/n`
So `lambda_B/lambda_G=n_B/n_G=3/1`
`lambda_B=3lambda_G`
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Knowledge Check

  • de-Broglie wavelength lambda is

    A
    proportional to mass
    B
    proportional to impulse
    C
    inversely proportional to impulse
    D
    independent to impulse

  • The De-Broglie wave length of electron in second exited state of hydrogen atom is

    A
    `5A`
    B
    `10A`
    C
    `100A`
    D
    `6.6A`

  • de-Broglie wavelength associated with an electron revolving in the n^(th) state of hydrogen atom is directily proportional to

    A
    n
    B
    `1/n`
    C
    `n^(2)`
    D
    `(1)/(n^(2))`

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