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An excited hydrogen atom returns to the ...

An excited hydrogen atom returns to the ground state . The wavelength of emitted photon is ` lambda` The principal quantum number of the excited state will be :

A

`[(lambdaR)/(lambdaR-1)]^(1//2)`

B

`[(lambdaR+1)/(lambdaR)]^(1//2)`

C

`[lambdaR(lambdaR+1)]^(1//2)`

D

`[(1)/(lambdaR(lambdaR+1))]^(1//2)`

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The correct Answer is:
A
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Knowledge Check

  • An excited hydrogen atom returns to the ground state . The wavelength of emittted photon is lambda The principal quantium number fo the excited state will be :

    A
    ` [(R+1)/(lambdaR)]^(2//2)`
    B
    `[lambdaR (lambdaR + 1) ]^(1//2)`
    C
    ` [(R+1)/(lambdaR-1)]^(2//2)`
    D
    `[1/(lambdaR (lambdaR+1))]^(1//2)`

  • Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength lambda . If R is the Rydberg constant, then the principal quatum number n of the excited state is

    A
    `sqrt((lambdaR)/(lambdaR-1))`
    B
    `sqrt((lambda)/(lambdaR-1))`
    C
    `sqrt((lambdaR^2)/(lambdaR-1))`
    D
    `sqrt((lambdaR)/(lambda-1))`

  • when a hydrogen atom is raised from the ground state to an excited state

    A
    `P.E.` increases and `K.E.` decreases
    B
    `P.E.` decreases and `K.E.` increases
    C
    Both kinetic energy and potential eenrgy increase
    D
    Both `K.E.` and `P.E`. Decrease
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