(1)
0. Potential energy is - 27.2 eV in second orbit of
He then calculate, double of total energy in first
excited state of Hydrogen atom :-
(1) - 13.6 eV
(2) - 54.4 eV
(3) - 6.8 eV
(4) - 27.2 eV
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Answer:
Potential Energy is -27.2 eV in second orbit of He
+
then calculate double of total energy in the first excited state of hydrogen atom - 6.8 eV.
Total energy is one half the potential energy.
Second orbit of He
+
has a total energy of −27.2 eV.
Hence, its total energy will be
2
−27.2
=−13.6eV
The energy is proportional to
n
2
Z
2
E
2,H
=E
2,He+
×
Z
He+
2
Z
H
2
=−13.6×
2
2
1
2
=−3.4 eV
The double of the total energy in the first excited state of hydrogen atom is 2×(−3.4eV)=−6.8 eV
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