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Multiple choice
In hydrogen the energy levels are
E
n
=
−
13.6
eV
/
n
2
E_n = -13.6\ \text{eV} / n^2
E
n
=
−
13.6
eV
/
n
2
. An electron drops from
n
=
4
n = 4
n
=
4
to
n
=
2
n = 2
n
=
2
, emitting a photon. Taking the photon energy as
E
initial
−
E
final
E_\text{initial} - E_\text{final}
E
initial
−
E
final
(
E
4
−
E
2
E_4 - E_2
E
4
−
E
2
), what is its magnitude (in eV)?
3.40
2.55
0.85
4.25
Check answer