Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

The document Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
All you need of JEE at this link: JEE

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 = Smax = 1

L = Lmax = 1 + 2 = 3 and J = L + S = 4

Then Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

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 = Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Hence g = 0. Thus

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

or Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 Hence Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Thus  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

Q.151. A certain atom is in the state in which S = 2, the total angular momentum Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev, and the magnetic moment is equal to zero. Write the spectral symbol of the corresponding term. 

Ans. From Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

we find J = 1. From the zero value of the magnetic moment we find

g = o

or  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

or          12 = L (L+ 1)

Hence L = 3 . The state is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

Q.152. An atom in the state 2P3/2  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 Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev is the total angular momentum vector of the atom then there is a magnetic moment

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

associated with it; here g is the Lande factor. In a magnetic field of induction Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev an energy

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Using Gaussian unit expression of Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev eig/gauss, B = 103 gauss

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev  erg sec and for the 2P3/2 state

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and             Ω = 1.17 x 1010 rad/s

The same formula is valid in MKS units also But μB = 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 2P1/2  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 Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevin a magnetic Geld of induction Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevis given by

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

In the present case, the maximum force arise when Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev is along the axis or close to it.

Then Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Here Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev  . The Lande factor g is for 2P1/2

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

The magnetic field is given by

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

or  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Thus  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Thus the maximum force is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

The force on a magnetic dipole of moment Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev due to this magnetic field is also tangential and has a magnitude

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

This force is nonvanishing only when the component of Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Then

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Now the maxim um value of Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Thus the force is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

Q.155. A narrow stream of vanadium atoms in the ground state 4F3/2  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 l = 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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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 Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

we get Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For vanadium atom in the ground state 4F3/2

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevusing other data, and substituting

we get Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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) 2F5/2; (c) 4P1/2

Ans. (a) The term 3P 0 does not split in weak magnetic field as it has zero total angular momentum.

(b) The term 2F5/2 will split into  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev sublevels. The shift in each sublevel is given by

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 where Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev and g is the Landi factor 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

(c) In this case for the 4D1/2 term 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Thus the splitting is 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Substitution gives  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

(b) For the 3F4 teim Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 and   Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where           Mj = - 4 to + 4 . Thus

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Substitution gives Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

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) 2D5/2 → 2P3/2; (c) D13P0; (d) 5l5   →  5H4

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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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

(b) The term 2D5/2 will split ot in 6 terms in accordance with the formula 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Ther term 2P3/2 will also split into 4 lines in accordance with the above formula with

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev and Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

It is seen that the Z eeman splitting is auomalous as g factors are different

(c)Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For the 5H4 term

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

We see that the splitting in the two levels given by Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev 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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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 Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Substitution gives Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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 μB .This must equal

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Thus Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

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 Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev then

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For resolution Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev of the instrument 

Thus 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Hence the minimum megnetic induction is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

Q.162. A spectral line caused by the transition 3D 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 3Pterm does not split The 3D1 term splits into 3 lines corresponding to the shift

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

with M= ± 1, 0. The interval between neighbouring components is then given by

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Hence

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Now for the 3D1 term

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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 2P3/2  and 2P1/2  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 2P3/2 is η = 50 times smaller than the natural splitting of the term 2P. 

Ans. (a) For the 2P3/2 term

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and the energy of the 2P3/2 sublevels will be

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevThus, between neighbouring sublevels.

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For the 2P1/2 terms

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

and the separation between the two sublevels into which the 2P1/2 term will split is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

The ratio of the two splittings is 2 : 1.

(b) The interval between neighbouring Zeem an sublevels o f the 2P3/2 term is Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev The energy separation between D1 and D2 lines is Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev  (this is the natural separation of the 2P them)

 Thus  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

or  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Substitution gives

B = 5-46kG

 

Q.164. Draw a diagram of permitted transitions between the terms 2P3/2 and 2S1/2 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 2P3/2 level g = 4 / 3 (see above) and the eneigies of sublevels are

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevfor the four sublevels

For the  Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev g = 2 (since L = 0) and 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Permitted transitions must have ΔMZ = 0, ± 1

Thus only the following transitions occur

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 These six lines are shown below

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 

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 

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

2. How many Zeeman components are observed in both directions if the spectral line is caused by the transition (a) 2P3/2 2S1/2 ; (b) 3P →  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

ΔMZ = 0, ± 1

In (2) the light is emitted along die direction of die field. Then the selection rules are

ΔMZ = ± 1

ΔMZ = 0 is forbidden.

(a) In the transition Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

This has been considered above. In (1) we get all the six lines shown in the problem above

In (2) the line corresponding to Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevis forbidden.

Then we get four lines

(b) 3P →  3S1

For the 3P2 level, Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

so the energies of the sublevels are

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For the 3S1 line, g = 2 and the energies of the sublevels are

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

where MZ = 1, ± 0 The lines are

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

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

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 and hence the shift of the component is the value of

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

subject to the selection rule Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 For 3P2,

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Thus Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

For the different transition we have the following table

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev    Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

There are 15 lines in all.

The lines farthest out are Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

 The splitting between them is the total splitting. It is

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Substitution gives Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

MCQs

,

Objective type Questions

,

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

,

ppt

,

Semester Notes

,

shortcuts and tricks

,

Important questions

,

Exam

,

practice quizzes

,

past year papers

,

Previous Year Questions with Solutions

,

Summary

,

Free

,

pdf

,

Sample Paper

,

Extra Questions

,

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

,

mock tests for examination

,

study material

,

Irodov Solutions: Properties of Atoms. Spectra- 4 Notes | EduRev

,

video lectures

,

Viva Questions

;