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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

  • Give no. of energy state orbital and electrons in N orbital .

    A
    4,12,32
    B
    4,16 ,30
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