Irodov Solutions: Properties of Atoms. Spectra- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

Q.149. An excited atom has the electronic configuration 1s²2s²2p³d 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 ²P_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 = 10³ gauss

  erg sec and for the ²P_3/2 state

and             Ω = 1.17 x 10¹⁰ rad/s

The same formula is valid in MKS units also But μ_B = 0.927 x 10 ^-23 A.m² , 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 ²P_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 ²P_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^-27N

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 ⁴F_3/2  is passed through a transverse strongly inhomogeneous magnetic field of length l₁ = 5.0 cm as in the Stern-Gerlach experiment. The beam splitting is observed on a screen located at a distance l₂  = 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 ⁴F_3/2

using other data, and substituting

we get 

This value differs from the answer given in the book by almost a factor of 10⁹. 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. 10⁹ 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) ³P₀; (b) ²F_5/2; (c) ⁴P_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 ²F_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 ⁴D_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) ¹D; (b) ³F₄.

Ans. (a) for the ¹D₂ term

and 

 Thus the splitting is 

Substitution gives  

(b) For the ³F₄ 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) ¹P → ¹S (b) ²D_5/2 → ²P_3/2; (c) ³ D₁ → ³P₀; (d) ⁵l₅   →  ⁵H₄? 

Ans. (a) The term ¹P₁splits into 3 lines with M_z = ± 1, 0 in accordance with the formula

where 

The term ¹S₀ does not split in weak magnetic field. Thus the transitions between ¹P₁ & ¹S₀ will result in 3 lines i.e. a normal Zeeman triplet.

(b) The term ²D_5/2 will split ot in 6 terms in accordance with the formula 

Ther term ²P_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 ³D₁ splits into 3 levels (g = 5 /2)

The term ³P₀ does not split. Thus the Z eeman spectrum is normal.

(d) For the 5I5 term 

For the ⁵H₄ 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 ¹F₃.

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 = = 1.0.10⁵  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 ³D₁ →  ³P₀ 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 = 1.32.10¹⁰ s ^-1 Find the magnetic field induction B at the point where the source is located.

Ans. The ³P₀ term does not split The ³D₁ 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 ³D₁ term

Substitution gives B = 3.00 kG. = 0.3 T.

Q.163. The wavelengths of the Na yellow doublet (²P → ²S) 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 ²P_3/2  and ²P_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 ²P_3/2 is η = 50 times smaller than the natural splitting of the term ²P. 

Ans. (a) For the ²P_3/2 term

and the energy of the ²P_3/2 sublevels will be

where  Thus, between neighbouring sublevels.

For the ²P_1/2 terms

and the separation between the two sublevels into which the ²P_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 ²P_3/2 term is  The energy separation between D₁ and D₂ lines is   (this is the natural separation of the ²P them)

 Thus  

or  

Substitution gives

B = 5-46kG

Q.164. Draw a diagram of permitted transitions between the terms ²P_3/2 and ²S_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 ²P_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) ²P_3/2 → ²S_1/2 ; (b) ³P₂  →  ³S₁? 

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) ³P₂  →  ³S₁

For the ³P₂ level, 

so the energies of the sublevels are

where 

For the ³S₁ 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 ³D₃ → → ³P₂  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 ³P₂,

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