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UCL PHYSICS AND ASTRONOMY 2224 Problem class 1
PHAS2224 ATOMIC AND MOLECULAR PHYSICSProblem Solving Tutorial 1
(201314)
Week 22 (20 24 January 2014)
1. A sample of hydrogen atoms is prepared in the 4p state. At
what wavelengths will lightsubsequently be observed? How would they
differ if the atoms were prepared in the 4sstate?(You may assume
that the transitions obey the selection rule nany, l= 1).
2. Calculate the wavenumber (in cm1) of the n= 4 to n= 3
transition of singly ionisedhelium (He+) assuming the mass of the
He nucleus to be infinite. The 4He nucleus is7294 time as massive
as an electron. By how much does this infinite mass effect shiftthe
transition wavenumber?(R= 109737.31 cm
1)
3. Positronium is an atomic-like species comprising an electron
and its anti-paricle the
positron. For states with principal quantum numbern compare the
orbital radius andenergy with those for atomic hydrogen with the
same n.
4. An atom has an electron orbital angular momentum of L = 2 and
an electron spinangular momentum ofS= 1. Calculate the possible
allowed values of the total electronangular momentumJ.
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Model Answers
1. 4p has n = 4 and l = 1: allowed transitions:
n= 1
n = 2
n = 3
n = 4
s (l = 0) d (l = 2) f (l = 3)p (l = 1)
This gives the following (note: important not to ignore cascade
transitions):n= 4 3, 4 2, 4 1, 3 2, 2 1 Wavelengths:
1
if=R
1
nf2
1
ni2
withR=109677.58 cm1 gives:1874 nm, 486.0 nm, 97.2 nm, 656.1 nm,
121.5 nm respectively.
4s has n = 4 and l = 0: allowed transitions:
n= 1
n = 2
n = 3
n = 4
s (l = 0) d (l = 2) f (l = 3)p (l = 1)
This gives the following : n = 4 3, 4 2, 3 2, 3 1, 2 1
Wavelengths: 1874 nm,486.0 nm, 656.1 nm, 102.5 nm, 121.5 nm
respectivelyie one sees Ly-instead of Ly-.
2.
= RZ2
1
nf2
1
ni2
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withR= 109737.31 cm1,Z= 2, nf= 3,ni= 4 gives = 21337.8 cm
1.
Finite nuclear mass effect. Need to scale R
R=
me
R
where the reduced mass
= mHememHe+me
=7294
7295me
now = 21334.9 cm1, a shift of 2.9 cm1.
3. Effect arises from reduced mass of positronium (Ps)
= mememe+me
=me
2
so reduced mass is half that of H (assuming infinite nuclear
mass).
rn=40h
2n2meZe2
=n2
a0
So radius of Ps orbit twice that of H atom with same n.
Energy
En= R
me
1
n2
so energy of Ps state is half that of H atom with same n.
4.J= |L S|, . . . L+Sin steps of one
So withL = 2 and S= 1, J= 1, 2, 3 ie 3 values.Gives levels:
3D1,
3D2, 3D3
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Derivation of the mass dependent formula for rn
Newtons 2nd law gives for circular motion:
Ze2
40r2 Coulomb
= mev
2
r Centripetal
. (1)
wherev is speed of electron,me mass of electron and r radius of
orbit.
Bohr postulated (1913) that stationary states, are those for
which the angular mo-mentum, L is quantised:
L= nh, h= h/2, n= 1, 2, 3, . . . (2)
i.e. L is an integral multiple of h; orbit must be an integral
number of de Brogliewavelengths of the electron.
SinceL = mevr
v= nh
mer. (3)
So, substituting forv in Eq. (1)
Ze2
40r =me
nh
mer
2. (4)
rn=40h
2n2
Ze2me. (5)
For principal quantum number n= 1,
rn=1=40h
2
Ze2me. (6)
ForZ= 1 (H) called the Bohr radius, a0, and rn= n2a0
