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For a doubly ionised Li-atom...

For a doubly ionised Li-atom

A

Angular momentum of an electron in `3rd` orbit is `(3h)/(2pi)`

B

energy of electron in `2nd` ecited state is `-13.6 eV`.

C

Speed of electron in `3rd` orbit is `(c)/(137)`, where c is speed of light

D

Kinetic energy of electron is 2nd excited state is half of the magnitude of the potential energy.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(A) `L = (nh)/(2pi)` , (B) `E = (-13.6Z^(2))/(n^(2))`
(C ) `v = (c )/(137)(Z)/(n)` , (D) `K = -(U)/(2)`
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Knowledge Check

  • As par Bohr model, the minimum energy (in eV ) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is

    A
    `1. 51`
    B
    `13.6`
    C
    `40.8`
    D
    `122.4`

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    A
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    B
    `(27)/(8)`
    C
    `(8)/(27)`
    D
    `(9)/(4)`

  • If lambda is the wavelength of hydrogen atom from the transition n=3 to n=1 ,then what is the wavelength for doubly ionised lithium ion for same transition?

    A
    `lambda/3`
    B
    `3lambda`
    C
    `lambda/9`
    D
    `9lambda`

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