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The electron in a hydrogen atom makes a ...

The electron in a hydrogen atom makes a transition `n_(1) rarr n_(2)` where `n_(1)` and `n_(2)` are the principal qunatum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is `1//27` of that in the final state. The possible values of `n_(1)` and `n_(2)` are

A

`n_(1) = 4, n_(2) = 2`

B

`n_(1) = 3, n_(2) = 1`

C

`n_(1) = 8, n_(2) = 1`

D

`n_(1) = 6, n_(2) = 3`

Text Solution

Verified by Experts

The correct Answer is:
B
`T prop n^(3)`
`T_(i)/T_(f)=1/27=(n/m)^(3) implies n/m=1/3`
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Knowledge Check

  • The electron in a hydrogen atom make a transtion n_(1) rarr n_(2) where n_(1) and n_(2) are the priocipal quantum number of the two states . Assume the Bohr model to be valid . The time period of the electron in the initial state is eight time that in the state . THe possible values of n_(1) and n_(2) are

    A
    `n_(1) = 4 , n_(2) = 2 `
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    `n_(1) = 8 , n_(2) = 2`
    C
    `n_(1) = 8 , n_(2) = 1 `
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    `n_(1) = 6 , n_(2) = 3`

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    ` n_1 =4, n_2 =2`
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