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An excited hydrogen atom emits a photon ...

An excited hydrogen atom emits a photon of wavelength `lambda` in returning to the ground state. If R is the Rydberg constant, then the quantum number n of the excited state is

A

`sqrt(lambda R)`

B

`sqrt(lambda R-1)`

C

`sqrt(lambda (lambdaR-1))`

D

`sqrt((lambda R)/(lambdaR-1))`

Text Solution

Verified by Experts

The correct Answer is:
D
`1/lambda =R_H(1/1^2 -1/n^2)`
`1-1/n^2 =1/(R_H lambda) =(lambdaR_H-1)/(lambdaR_H)`
`n^2 =(lambdaR_H)/(lambdaR_H-1)`
`:. n = sqrt((lambdaR_H)/(lambdaR_H-1))`
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Knowledge Check

  • An excited hydrogen atom emits a photon of wavelength lambda in returning to the ground state. If 'R' is the Rydberg's constant, then the quantum number 'n' of the excited state is:

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

  • 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))`

  • A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is 13.6 eV, the energy (E_t) and quantum number (n) of the resulting state are respectively,

    A
    `E_t = -13.6eV , n =1`
    B
    `E_t =- 6.0 eV , n =3`
    C
    `E_t = -6.0 eV , n=2`
    D
    `E_t = - 13.6 eV , n=2`

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