Question

the energy of a hydrogen atom in the ground state is -13.6 calculate energy of he+ ion in first excited state

Answer 1

This is only known exactly for the hydrogen-like atoms. Otherwise, it is done experimentally via photoelectron spectroscopy.
For hydrogen-like atoms, i.e.
H
,
He
+
,
Li
2
+
, etc., the energy levels are given by:
E
n
=
−
Z
2
⋅
13.61 eV
n
2
where
Z
is the atomic number and
n
is the quantum level.
So for
He
+
, the first excited state energy level would be the
1
s
0
2
p
1
configuration:
E
2
=
−
2
2
⋅
13.61 eV
2
2
=
−
13.61 eV
And its ground state energy would be:
E
1
=
−
2
2
⋅
13.61 eV
1
2
=
−
54.44 eV
So, its first excited state lies
40.83 eV
above its ground state. That matches the electronic energy level difference here from NIST:

https://www.physics.nist.gov/

329179
cm
−
1
×
2.998
×
10
10
cm
s
×
6.626
×
10
−
34
J
⋅
s
×
1 eV
1.602
×
10
−
19
J
=
40.82 eV
≈
40.83 eV
−−−−−−−−