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524
Atomic Physics
Section 42.4
The Quantum Model of the Hydrogen Atom
P42.16
The reduced mass of positronium is
less
than hydrogen, so the photon energy will be
less
for
positronium than for hydrogen. This means that the wavelength of the emitted photon will be
longer
than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge
than hydrogen, so its transition energy will be
larger
, corresponding to a wavelength
shorter
than
656.3 nm.
All the factors in the given equation are constant for this problem except for the reduced mass
and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the
transition can be found simply from the ratio of mass and charge variables.
For hydrogen,
µ
=
+
≈
mm
m
pe
e
.
The photon energy is
∆
EE E
=−
32
.
Its wavelength is
λ
=
656 3
. nm,
where
==
c
f
hc
E
∆
.
(a)
For positronium,
=
+
=
m
ee
e
2
so the energy of each level is one half as large as in hydrogen, which we could call
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Energy, Mass, Photon

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