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The electron in a hydrogen atom jumps ba...

The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of wavelength `lambda_(0) = (16)/(15R)`, where R is Rydbergs's constant. In place of emitting one photon, the electron could come back to ground state by

A

Emitting 3 photons of wavelengths `lambda_(1),lambda_(2)` and `lambda_(3)` such that `(1)/(lambda_(1)) +(1)/(lambda_(2))+(1)/(lambda_(3)) = (15R)/(16)`

B

Emitting 2 photons of wavelength `lambda_(1)` and `lambda_(2)` such that `(1)/(lambda_(1))+(1)/(lambda_(2)) = (15R)/(16)`

C

Emitting 2 photons of wavelength `lambda_(1)` and `lambda_(2)` such that `lambda_(1)+lambda_(2) = (16)/(15R)`

D

Emitting 3 photons of wavelength `lambda_(1),lambda_(2)` and `lambda_(3)` such that `lambda_(1)+lambda_(2)+lambda_(3) = (16)/(15R)`

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Knowledge Check

  • 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 hydrogen atom ia in excited state of principal quantum number n . It emits a photon of wavelength lambda when it returnesto the ground state. The value of n is

    A
    `sqrt(lambda R (lambda R - 1))`
    B
    `sqrt((lambda (R - 1))/(lambda R))`
    C
    `sqrt((lambda R)/(lambda R - 1))`
    D
    `sqrt lambda (R - 1)`
  • An electron in a hydrogen atom makes a transition from first excited state to ground state . The equivelent currect due to circulating electron :

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    increase 2 time
    B
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