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de-Broglie wavelength of an electron in ...

de-Broglie wavelength of an electron in the nth Bohr orbit is `lambda_(n)` and the angular momentum is `J_(n)` then

A

`J_n alpha lamda_n`

B

`lamda_n alpha 1//J_n`

C

`J_n alpha sqrtlamda_n`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
de-Broglie wavelength
`lamda = h/p = h/(mv) implies lamda alpha 1//v`
Now, v` alpha` Z/n or v `alpha ` 1/n
[By Bohr atom relation]
Hence, `lamda`n `alpha` n ... .. (i)
i.e., wavelength in nth orbit `lamda`n is
proportional to n (quantum number)
Now `J_n` = n`h/2pi`...(ii)
[ `J_n` = angular momentum of electron in nth orbit]
i.e. `J_n alpha` n
Hence, from Eqs. (i) and (ii),
`J_n alpha lamda_n`
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Knowledge Check

  • 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

  • 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 first Bohr orbit is equal to

    A
    the circumference of first orbit
    B
    (1/2) `xx` (circumference of first orbit )
    C
    (1/4) `xx` (circumference of first orbit )
    D
    (3/4) `xx` (the circumference of first orbit )

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