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The ionization energy of the hydrogen ...

The ionization energy of the hydrogen atom is given to be 13.6 eV .A photon falls on hydrogen atom
which is initially in the ground state and excited in to the(n=4) state.
(i) Show this transition in the energy- level digram &
(ii) calcuted the wavelenght of the photon.

Text Solution

Verified by Experts

The correct Answer is:
`973.5Å`
`DeltaE=13.6[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=(hc)/(lambda)`
`lambda=973.5Å`
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Knowledge Check

  • The ionization potential of the hydrogen atom is 13.6 V. The energy needed to ionize a hydrogen atom which is in its first excited state is about

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    D
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  • A photon of energy 12.09 eV is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of excited state is

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    B
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