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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.
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.
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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 =E4−E1=(−0.85)−(−13.6)=12.75 eV≈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 =E4−E1=(−0.85)−(−13.6)=12.75 eV≈12.8 eV
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