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The ratio of the speed of the electrons ...

The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

A

`1//2`

B

`2//137`

C

`1//137`

D

`1//237`

Text Solution

Verified by Experts

The correct Answer is:
C

Let the speed of electron in `n^(th)` orbit of atom be `v_(n)`.
We know, `v_(n) = (Ze^(2))/(2epsilon_(0)nh)`
For ground state, n = 1
`v_(1) = e^(2)/(2epsilon_(0) xx 1 xx h)`
`v_(1)/c = (e^(2)/(2 epsilon_(0)h) (1/c))`
`(v_(1)/c) = ((1.6 xx 10^(-19)) xx (1.6 xx 10^(-19)))/(2 xx 8.85 xx 10^(-12) xx 6.63 xx 10^(-34) xx 3 xx 10^(8))`
`(v_(1))/c = 1/137`
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Knowledge Check

  • The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where e, h and c have their usual meaning in cgs system)

    A
    `2 pi h // e^(2)`
    B
    `e r^(2) h// 2 pi c`
    C
    `e ^(2) c// 2 pi h`
    D
    `2 pi e^(2)// h c`
  • The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where e, h and c have their usual meanings)

    A
    `2pihc//e^(2)`
    B
    `e^(2)h//2pic`
    C
    `e^(2)c//2pih`
    D
    `2pie^(2)//hc`
  • If omega the speed of electron in the nth orbit hydrogen atom, then

    A
    `omega propn^(1//2)`
    B
    `omegaprop(1)/(n)`
    C
    `omega prop(1)/(n^(2))`
    D
    `omegaprop(1)/(n^(3))`
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