Q.149. An excited atom has the electronic configuration 1s^{2}2s^{2}2p^{3}d being in the state with the greatest possible total angular momentum. Find the magnetic moment of the atom in that state.
Ans. To get the greatest possible angular momentum we must have S = S_{max} = 1
L = L_{max} = 1 + 2 = 3 and J = L + S = 4
Then
and
Q.150. Find the total angular momentum of an atom in the state with S = 3/2 and L = 2 if its magnetic moment is known to be equal to zero.
Ans. Since μ = 0 we must have either J = 0 or g = 0. But J = 0 is incompatible with L = 2 and S = Hence g = 0. Thus
or
Hence
Thus
Q.151. A certain atom is in the state in which S = 2, the total angular momentum , and the magnetic moment is equal to zero. Write the spectral symbol of the corresponding term.
Ans. From
we find J = 1. From the zero value of the magnetic moment we find
g = o
or
or 12 = L (L+ 1)
Hence L = 3 . The state is
Q.152. An atom in the state ^{2}P_{3/2 } is located in the external magnetic field of induction B = 1.0 kG. In terms of the vector model find the angular precession velocity of the total angular momentum of that atom.
Ans. If is the total angular momentum vector of the atom then there is a magnetic moment
associated with it; here g is the Lande factor. In a magnetic field of induction an energy
is associated with it. This interaction term corresponds to a presession of the angular momentum vector because if leads to an equation of motion of the angular momentum vector of the form
where
Using Gaussian unit expression of eig/gauss, B = 10^{3} gauss
erg sec and for the^{ 2}P_{3/2} state
and Ω = 1.17 x 10^{10} rad/s
The same formula is valid in MKS units also But μ_{B} = 0.927 x 10^{ 23} A.m^{2} , B = 10^{1 }T and :h = 1.054 x 10^{34 }Joule sec. The answer is the same.
Q.153. An atom in the state ^{2}P_{1/2 } is located on the axis of a loop of radius r = 5 cm carrying a current I = 10 A. The distance between the atom and the centre of the loop is equal to the radius of the latter. How great may be the maximum force that the magnetic field of that current exerts on the atom?
The force on an atom with magnetic moment in a magnetic Geld of induction is given by
In the present case, the maximum force arise when is along the axis or close to it.
Then
Here . The Lande factor g is for ^{2}P_{1/2}
and
The magnetic field is given by
or
Thus
Thus the maximum force is
Substitution gives (using data in M K S units)
F = 4.1 x 10^{27}N
Q.154. A hydrogen atom in the normal state is located at a distance r = 2.5 cm from a long straight conductor carrying a current I = 10 A. Find the force acting on the atom.
Ans.The magnetic field at a distance r from a long current carrying wire is mostly tangential and given by
The force on a magnetic dipole of moment due to this magnetic field is also tangential and has a magnitude
This force is nonvanishing only when the component of Then
Now the maxim um value of Thus the force is
Q.155. A narrow stream of vanadium atoms in the ground state ^{4}F_{3/2} is passed through a transverse strongly inhomogeneous magnetic field of length l_{1} = 5.0 cm as in the SternGerlach experiment. The beam splitting is observed on a screen located at a distance l_{2 } = 15 cm from the magnet. The kinetic energy of the atoms is T = 22 MeV. At what value of the gradient of the magnetic field induction B is the distance between the extreme components of the split beam on the screen equal to δ = 2.0 mm?
Ans. In the homogeneous magnetic field the atom experinces a force
Depending on the sign of /, this can be either upward or downward. Suppose the latter is true. The atom then traverses first along a parabola inside the field and, once outside, in a straight line. The total distance between extreme lines on the screen will be
Here mv is the mass of the vanadium atom. (The first term is the displacement within the field and the second term is the displacement due to the transverse velocity acquired in the magnetic field).
Thus using
we get
For vanadium atom in the ground state ^{4}F_{3/2}
using other data, and substituting
we get
This value differs from the answer given in the book by almost a factor of 10^{9}. For neutral atoms in stem Gerlach experiments, the value T = 22MeV is much too laige. A more appropriate value will be T = 22 meV i.e. 10^{9} times smaller. Then one gets the right answer.
Q.156. Into what number of sublevels are the following terms split in a weak magnetic field: (a) ^{3}P_{0}; (b) ^{2}F_{5/2}; (c) ^{4}P_{1/2}?
Ans. (a) The term 3P_{ 0} does not split in weak magnetic field as it has zero total angular momentum.
(b) The term ^{2}F_{5/2} will split into sublevels. The shift in each sublevel is given by
where and g is the Landi factor
(c) In this case for the ^{4}D_{1/2} term
Thus the energy differences vanish and the level does not split.
Q.157. An atom is located in a magnetic field of induction B = 2.50 kG. Find the value of the total splitting of the following terms (expressed in eV units): (a) ^{1}D; (b) ^{3}F_{4}.
Ans. (a) for the ^{1}D_{2} term
and
Thus the splitting is
Substitution gives
(b) For the ^{3}F_{4} teim
and
where Mj =  4 to + 4 . Thus
Substitution gives
Q.158. What kind of Zeeman effect, normal or anomalous, is observed in a weak magnetic field in the case of spectral lines caused by the following transitions: (a) ^{1}P → ^{1}S (b) ^{2}D_{5/2} → ^{2}P_{3/2}; (c) ^{3 }D_{1} → ^{3}P_{0}; (d) ^{5}l_{5} → ^{5}H_{4}?
Ans. (a) The term ^{1}P_{1}splits into 3 lines with M_{z} = ± 1, 0 in accordance with the formula
where
The term ^{1}S_{0} does not split in weak magnetic field. Thus the transitions between ^{1}P_{1} & ^{1}S_{0 }will result in 3 lines i.e. a normal Zeeman triplet.
(b) The term ^{2}D_{5/2} will split ot in 6 terms in accordance with the formula
Ther term ^{2}P_{3/2 }will also split into 4 lines in accordance with the above formula with
and
It is seen that the Z eeman splitting is auomalous as g factors are different
(c)
The term ^{3}D_{1} splits into 3 levels (g = 5 /2)
The term ^{3}P_{0} does not split. Thus the Z eeman spectrum is normal.
(d) For the 5I5 term
For the ^{5}H_{4} term
We see that the splitting in the two levels given by is the same though the number of levels is different (11 and 9). It is then easy to see that only the lines with following energies occur
The Z eeman pattern is normal
Q.159. Determine the spectral symbol of an atomic singlet term if the total splitting of that term in a weak magnetic field of induction B = 3.0 kG amounts to ΔE = 104μeV
Ans. For a singlet term S = 0, L = J, g = 1
Then the total splitting is
Substitution gives
The term is ^{1}F_{3}.
Q.160. It is known that a spectral line λ = 612 nm of an atom is caused by a transition between singlet terms. Calculate the interval Δλ between the extreme components of that line in the magnetic field with induction B = 10.0 kG.
Ans. As the spectral line is caused by transition between singlet terms, the Z eeman effect will be normal (since g = 1 for both terms). The energy difference between extreme components of the line will be 2 μ_{B }B .This must equal
Thus
Q.161. Find the minimum magnitude of the magnetic field induction B at which a spectral instrument with resolving power λ/δλ = = 1.0.10^{5} is capable of resolving the components of the spectral line λ = 536 nm caused by a transition between singlet terms. The observation line is at right angles to the magnetic field direction.
Ans. From the previous problem, if the components are then
For resolution of the instrument
Thus
Hence the minimum megnetic induction is
Q.162. A spectral line caused by the transition^{ 3}D_{1 }→ ^{ 3}P_{0} experiences the Zeeman splitting in a weak magnetic field. When observed at right angles to the magnetic field direction, the interval between the neighbouring components of the split line is Δω = 1.32.10^{10} s ^{1} Find the magnetic field induction B at the point where the source is located.
Ans. The ^{3}P_{0 }term does not split The ^{3}D_{1} term splits into 3 lines corresponding to the shift
with M_{z }= ± 1, 0. The interval between neighbouring components is then given by
Hence
Now for the ^{3}D_{1} term
Substitution gives B = 3.00 kG. = 0.3 T.
Q.163. The wavelengths of the Na yellow doublet (^{2}P → ^{2}S) are equal to 589.59 and 589.00 nm. Find:
(a) the ratio of the intervals between neighbouring sublevels of the Zeeman splitting of the terms ^{2}P_{3/2 } and ^{2}P_{1/2 } in a weak magnetic field;
(b) the magnetic field induction B at which the interval between neighbouring sublevels of the Zeeman splitting of the term ^{2}P_{3/2 }is η = 50 times smaller than the natural splitting of the term ^{2}P.
Ans. (a) For the ^{2}P_{3/2 }term
and the energy of the ^{2}P_{3/2} sublevels will be
where Thus, between neighbouring sublevels.
For the ^{2}P_{1/2} terms
and the separation between the two sublevels into which the ^{2}P_{1/2} term will split is
The ratio of the two splittings is 2 : 1.
(b) The interval between neighbouring Zeem an sublevels o f the ^{2}P_{3/2} term is The energy separation between D_{1} and D_{2} lines is (this is the natural separation of the ^{2}P them)
Thus
or
Substitution gives
B = 546kG
Q.164. Draw a diagram of permitted transitions between the terms^{ 2}P_{3/2} and ^{2}S_{1/2 }in a weak magnetic field. Find the displacements (in rad/s units) of Zeeman components of that line in a magnetic field B = 4.5 kG.
Ans. for the ^{2}P_{3/2} level g = 4 / 3 (see above) and the eneigies of sublevels are
where for the four sublevels
For the g = 2 (since L = 0) and
where
Permitted transitions must have ΔM_{Z =} 0, ± 1
Thus only the following transitions occur
These six lines are shown below
Q.165. The same spectral line undergoing anomalous Zeeman splitting is observed in direction 1 and, after reflection from the mirror M (Fig. 6.9), in direction
2. How many Zeeman components are observed in both directions if the spectral line is caused by the transition (a) ^{2}P_{3/2 }→ ^{2}S_{1/2} ; (b) ^{3}P_{2 } → ^{3}S_{1}?
Ans.The difference arises because of different selection rules in the two cases. In (1) the line is emitted perpendicular to the field. The selection rules are then
ΔM_{Z =} 0, ± 1
In (2) the light is emitted along die direction of die field. Then the selection rules are
ΔM_{Z =} ± 1
ΔM_{Z =} 0 is forbidden.
(a) In the transition
This has been considered above. In (1) we get all the six lines shown in the problem above
In (2) the line corresponding to is forbidden.
Then we get four lines
(b) ^{3}P_{2 } → ^{3}S_{1}
For the ^{3}P_{2} level,
so the energies of the sublevels are
where
For the ^{3}S_{1} line, g = 2 and the energies of the sublevels are
where M_{Z =} 1, ± 0 The lines are
All energy differences are unequal because die two g values are unequal. There are then nine lines if viewed along (1) and Six lines if viewed along (2).
Q.166. Calculate the total splitting Ae) of the spectral line ^{3}D_{3} → → ^{3}P_{2} in a weak magnetic field with induction B = 3.4 kG.
Ans. For the two levels
and hence the shift of the component is the value of
subject to the selection rule
For ^{3}P_{2},
Thus
For the different transition we have the following table
There are 15 lines in all.
The lines farthest out are
The splitting between them is the total splitting. It is
Substitution gives
Irodov Solutions: Molecules and Crystals 1 Doc  10 pages 
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