22 2 n forntheenergyiszero

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Unformatted text preview: ⎛ n ⎞ m⎜ ⎟ ⎝ mr ⎠ Ze2 = r r n 2 2 Ze2 n22 = ⇒r= 2 mr r Zme2 2 a0 = (Bohr radius) 2 me Ioniza,on
energy
(I)
 •  When
the
distance
between
the
nucelus
and
 the
electron
is
very
large,
the
electron
is
 dissociated
from
the
proton(s),
and
the
 interac,on
energy
between
the
electron
and
 the
proton(s)
must
be
zero
 e
 p
 1 2 Ze2 E = mv − 2 r using the centrifugal/electrostatic force relation mv 2 Ze2 Ze2 2 = 2 → mv = r r r Ze2 E=− →0 r→∞ 2r Ioniza,on
energy
(II)
 •  Using
the
expression
for
the
distance
r
we
can
 write
the
energy
as
a
func,on
of
the
quantum
 number
n
 n22 r= Zme2 1 Ze2 E=− 2r 1 Z 2 e4 m E=− n = 1, 2, 3, ... 22 2 n •  For
n‐>∞
the
energy
is
zero
 •  The
lowest
energy
(most
stable
is
when
n=1).
 The
dissocia,on
energy
is
therefore
 ⎛ 1 Z 2 e4 m ⎞ 1 Z 2 e4 m Ionization energy=En = ∞ − En =1 = 0 − ⎜ − 2 ⎟ = 2 2 ⎝2 ⎠ Spectrum
of
hydrogen
atom
 •  A
transi,on
between
any
orbit‐states
ni
and
nf
 of
the
electron
in
a
hydrogen‐like
atoms
has
 an
energy
difference
 1 Z 2 e4 m ⎛ 1 1⎞ ΔE = E f − Ei = ⎜ n2 −...
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This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas at Austin.

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