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A hydrogen atom in a state of binding en...

A hydrogen atom in a state of binding energy 0.85 eV makes a transtition to a state of excitation energy of 10.2 eV.
(i) What is the initial state of hydrogen atom?
(ii) What is the final state of hydrogen atom ?
(iii) What is the wavelength of the photon emitted ?

Text Solution

Verified by Experts

Let `n_(2)` be the final excitation state of the electron, Since excitation energy is always measured with respect to the ground state, therefore
`DeltaE=13.6[1-(1)/(n_(2)^(2))]` here `DeltaE=10.2eV`, therefore,
`10.2=13.6[1-(1)/(n_(2)^(2))]` or `n_(2)=2` Thus, the electron jumps from `n_(1)=4` to `n_(2)=2`.
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Knowledge Check

  • When a hydrogen atom is raised from the ground state to fifth state :

    A
    both KE and PE increase
    B
    both KE and PE decrease
    C
    PE increase and KE decrease
    D
    PE decrease and KE increase

  • The ratio of the energies of hydrogen atom in the first to second exited state in

    A
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    B
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    C
    `4:9`
    D
    `9:4`

  • The radius of the hydrogen atom in the ground state is of the order of

    A
    `10^(-4) cm`
    B
    `10^(-5)cm`
    C
    `10^(-7)cm`
    D
    `10^(-8)cm`

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