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Question

Calculate the energy gap between the first energy state and the third excited state of a hydrogen like atom, given that the energy of the first excited state is -3.4 eV.

A
3.4 eV
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B
12 eV
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C
12.8 eV
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D
1.51 eV
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Solution

The correct option is C 12.8 eV
We know that the energy of nth level in a Hydrogen like atom is given by =13.6 (Z2n2)
According to the question, energy of the first excited state i.e. n = 2 is -3.4 eV.
So, 13.6Z2n2=3.4; where, n = 2
=13.6×Z24=3.4
Z=1

Energy of the first state,
=(E1)=13.6(1)2(1)2=13.6eV

Energy of the third excited state i.e., n=4(E4)
=13.6(1)2(4)2=0.85 eV

Hence, the energy gap between the third excited state and the first energy state =E4E1=(0.85)(13.6)=12.75 eV12.8 eV

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