Answered

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A hydrogen atom in an initial state having an ionization energy (the energy required to remove the electron from the hydrogen atom) of 0.85 eV makes a direct transition to a final state with an excitation energy (the difference in energy between the state and the ground state) of 10.2 eV.

Required:
a. Calculate the energy (in eV) of the emitted photon during the transition described above.
b. Consider all possible transitions from the initial state and final state of the hydrogen atom described above.

Sagot :

Answer:

Explanation:

a )

Energy of electron in fourth orbit of hydrogen atom = - 13.6 eV/ 4²

= -  .85 eV .

Energy of ground state  ( n = 1 ) = - 13.6 eV

energy of orbit n = 2

= 13.4 eV / 2² = -3.4 eV

difference = - 13.6 - ( - 3.4 eV)

= 10.2 eV

So transition is from 4 th orbit to 2 nd orbit

Energy difference = - 0.85 eV - ( - 3.4 eV )

= 2.55 eV

Energy of emitted photon = 2.55 eV .

b )

Possible transitions

1 ) from n = 4 to n = 3

2 ) from n = 3 to n = 2

3 ) from n = 4 to n = 2 .

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