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The number of revolutions per second mad...

The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of `3`:

A

`6.57xx10^(15)`

B

`6.57xx10^(13)`

C

`6.57xx10^(11)`

D

`6.57xx10^(14)`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The energy of an electron in the nth Bohr orbit of hydrogen atom is

    A
    `-(13.6)/(n^(4))`eV
    B
    `-(13.6)/(n^(3))`eV
    C
    `-(13.6)/(n^(2)) ` eV
    D
    `-(13.6)/(n) ` eV

  • The number of revolutions done by an electron 'e' in one second in the first orbit of hydrogen atom is

    A
    `6.57 xx 10^(15)`
    B
    `6.57 xx 10^(13)`
    C
    `1000`
    D
    `6.57 xx 10^(14)`.

  • The number of revolutions per second traced by an electron in the 2nd bohr orbit of Li^(+2) will be

    A
    `6.15 xx 10^(16)` rps
    B
    `7.37 xx 10^(15)` rps
    C
    `1.53 xx 10^(11)` rps
    D
    `8.92 xx 10^(16)` rps

    • Similar Questions

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    Calculate the frequency of revolution of electron in the first Bohr orbit of hydrogen atom, if radius of first Bohr orbit is 0.5Å and velocity of electron in the first orbit is 2.24xx10^6m//s .

    The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is

    The energy of an electron in second Bohr orbit of hydrogen atom is :

    The current in the first orbit of Bohr's hydrogen atom is

    The number of revolutions made by electron in Bohr's 2nd orbit of hydrogen atom is

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