Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

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

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 Q.167. Determine the angular rotation velocity of an S2 molecule promoted to the first excited rotational level if the distance between its nuclei is d = 189 pm. 

Ans. In the first excited rotational level J = 1

so Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Now Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where m is the mass of the mole cub and ri is the distance of the atom from the axis.

Thus  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

 Q.168. For an HCl molecule find the rotational quantum numbers of two neighbouring levels whose energies differ by 7.86 meV. The nuclei of the molecule are separated by the distance of 127.5 pm.

Ans. The axis of rotation passes through die centre of mass of the HCl molecule. The distances of the two atoms from the centre of mass are

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus I - moment of inertia about the axis

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

The energy difference ^between two neighbouring levels whose quantum numbers are J & J - 1 is

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Hence J = 3 and the levels have quantum numbers 2 & 3,

 

 Q.169. Find the angular momentum of an oxygen molecule whose rotational energy is E = 2.16 meV and the distance between the nuclei is d = 121 pm.

Ans. The angular momentum is

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Now Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev (m = mass of Omolecule) = 1.9584 x 10-39 gm c2

So Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(This corresponds to J = 3)

 

 Q.170. Show that the frequency intervals between the neighbouring spectral lines of a true rotational spectrum of a diatomic molecule are equal. Find the moment of inertia and the distance between the nuclei of a CH molecule if the intervals between the neighbouring lines of the true rotational spectrum of these molecules are equal to Δω = 5.47.1012 s-1.

Ans. From   Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

and the selection rule ΔJ = 1 or J → J - 1 for a pure rotational spectrum we get

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus transition lines are equispaced in frequency  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

In the case of CH molecule

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Also  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

so      d = 1.117 x 10-8 cm = 111.7 pm

 

Q.171. For an HF molecule find the number of rotational levels located between the zeroth and first excited vibrational levels assuming rotational states to be independent of vibrational ones. The natural vibration frequency of this molecule is equal to 7.79.1014  rad/s, and the distance between the nuclei is 91.7 pm.

Ans. if the vibrational frequency is to0 the excitation energy of the first vibrational level will be Irodov Solutions: Molecules and Crystals- 1 Notes | EduRevo. Thus if there are J rotational levels contained in the band between the ground state and the first vibrational excitation, then

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where as stated in the problem we have ignored any coupling between the two. For HF molecule

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Then Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

For J = 14 , J (J + 1 ) = 210 . For J - 13, J (J + 1) = 182. Thus there lie 13 levels between the ground state and the first vibrational excitation.

 

Q.172. Evaluate how many lines there are in a true rotational spectrum of CO molecules whose natural vibration frequency is ω = 4.09.1014  s-1  and moment of inertia I = 1.44.10-39g•cm2

Ans. We proceed as above. Calculating Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev we get

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Now this must equal  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Taking the square root we get J = 33 .

 

Q.173. Find the number of rotational levels per unit energy interval, dN/dE, for a diatomic molecule as a function of rotational energy E. Calculate that magnitude for an iodine molecule in the state with rotational quantum number J = 10. The distance between the nuclei of that molecule is equal to 267 pm. 

Ans. From the formula

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev  we get  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

or Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Hence Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

writing Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

we find   Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

The quantity Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev For laige E it is 

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

For an iodine molecule

I = mId2/ 2 = 7*57 x 10-38 gm cm2

Thus for J = 10

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRevIrodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Substitution gives

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.174. Find the ratio of energies required to excite a diatomic molecule to the first vibrational and to the first rotational level. Calculate that ratio for the following molecules:

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Here ω is the natural vibration frequency of a molecule, d is the distance between nuclei. 

 

Ans. For the first rotational level

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

for the first vibrational level Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Here ω = frequency of vibration. Now

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(a) For H2 molecule I = 4-58 x 10-41 gm cm2 and  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(b) For HI molecule,

I = 4.2 47 x 10-4Ogmcm2 and Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(c) For I2 molecule 

I = 7.57 x 10-38gmcm2 andIrodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.175. The natural vibration frequency of a hydrogen molecule is equal to 8.25.1014  s-1, the distance between the nuclei is 74 pm, Find the ratio of the number of these molecules at the first excited vibrational level (v = 1) to the number of molecules at the first excited rotational level (J = 1) at a temperature T = 875 K. It should be remembered that the degeneracy of rotational levels is equal to 2J + 1. 

Ans. The energy of the molecule in the first rotational level will be  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev.The ratio of tire number of molecules at tire first excited vibrational level to the number of molecules at the first excited rotational level is

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where             B = h /2I

For the hydrogen molecule Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Substitution gives 3.04 x 10-4

 

Q.176. Derive Eq. (6.4c), making use of the Boltzmann , distribution. From Eq. (6.4c) obtain the expression for molar vibration heat capacity Cv vib of diatomic gas. Calculate Cv vib for Cl2  gas at the temperature 300 K. The natural vibration frequency of these molecules is equal to 1.064 • 1014 s- 1 

Ans. By definition

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus for one gm mole of diatomic gas

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where R = Nk is the gas constant

In the present case  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

and  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.177. In the middle of the rotation--vibration band of emission spectrum of HCl molecule, where the "zeroth" line is forbidden by the selection rules, the interval between neighbouring lines is Δω = = 0.79.1013s-1. Calculate the distance between the nuclei of an HCl molecule.

Ans. In the rotation vibration band the main transition is due to change in vibrational quantum number v → v - 1. Together with this rotational quantum number may change. The “Zcroeth line” 0 →  0 is forbidden in this case so the neighbouring lines arise due to 1 →  0 or 0 → 1 in the rotational quantum number. Now

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Hence  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

so   Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Substitution gives d = 0.128 nm .

 

Q.178. Calculate the wavelengths of the red and violet satellites, closest to the fixed line, in the vibration spectrum of Raman scattering by F2 molecules if the incident light wavelength is equal to λ0 = 404.7 nm and the natural vibration frequency of the molecule is ω = 2.15.1014  s-1

Ans. If λR = wavelength of the red satellite

and λv = wavelength of the violet satellite

then Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

and  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Substitution gives

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

The two formulas can be combined to give

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.179. Find the natural vibration frequency and the quasielastic force coefficient of an S2  molecule if the wavelengths of the red and violet satellites, closest to the fixed line, in the vibration spectrum of Raman scattering are equal to 346.6 and 330.0 nm.

Ans. As in the previous problem

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev = 1.368 x 1014 rad/s

The force constant x is defined by

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where μ= reduced mass of the S2 molecule.

Substitution gives x = 5.01N/cm 

 

Q.180. Find the ratio of intensities of the violet and red satellites, closest to the fixed line, in the vibration spectrum of Raman scattering by Cl2 molecules at a temperature T = 300 K if the natural vibration frequency of these molecules is ω = 1.06.1014 s-1   . By what factor will this ratio change if the temperature is doubled? 

 

Ans. The violet satellite arises from the transition 1→0 in the vibrational state of the scattering molecule while the red satellite arises from the transition 0 → 1. The intensities of these two transitions are in the ratio of initial populations of the two states i.e. in the ratio

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

If the temperature is doubled, the rato increases to 0.259, an increase of 3.9 times.

 

Q.181. Consider the possible vibration modes in the following linear molecules: 

(a) CO(O — C—O); (b) C2H2 (H—C —C—H). 

 

Ans. (a) CO(O - C - O)

The molecule has 9 degrees of freedom 3 for each atom. This means that it can have up to nine frequencies. 3 degrees of freedom correspond to rigid translation, the frequency associated with this is zero as the potential energy of the system can not change under rigid translation. The P.E. will not change under rotations about axes passing through the C-atom and perpendicular to the O - C - O line. Thus there can be at most four non zero frequencies. We must look for modes different from the above.

One mode is Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Another mode is Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

These are the only collinear modes.

A third mode is doubly degenerate : Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(vibration in & Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev to the plane of paper).

(b) C2H2 ( H - C - C - H )

There are 4 x 3 - 3 - 2 = 7 different vibrations. There are three collinear modes.

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Two other doubly degenerate frequencies are

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

together with their counterparts in the plane Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev to the paper.

 

 

Q.182. Find the number of natural transverse vibrations of a string of length l in the frequency interval from ω to ω + dω if the propagation velocity of vibrations is equal to v. All vibrations are supposed to occur in one plane. 

 

Ans. Suppose the string is stretched along the x axis from x = 0 to x = l with the end points fixed. Suppose y(x, t) is the transverse displacement of the element at x at time t. Then y (x, t) obeys

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

We look for a stationary wave solution of this equation

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where A & δ are constants.. In this from y = 0 at x = 0. The further condition

y = 0 at x = l

implies Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

or  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

N is the number of modes of frequency ≤ ω.

Thus  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.183. There is a square membrane of area S. Find the number of natural vibrations perpendicular to its plane in the frequency interval from ω to ω + dω if the propagation velocity of vibrations is equal to v. 

Ans. Let Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev be the displacement of the element at (x , y) at time t. Then it obeys the equation

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

We look for a solution in the form

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Then Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

we write this as

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Here n, m > 0 . Each pair (n , m) determines a mode. The total mumber of modes whose frequency is ≤ ω is the area of the quadrant of a circle of radius Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev i.e.

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Then Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where S = I2 is the area of the membrane.

 

Q.184. Find the number of natural transverse vibrations of a rightangled parallelepiped of volume V in the frequency interval from ω to ω + dω if the propagation velocity of vibrations is equal to v.

 

Ans. For transverse vibrations of a 3-dimensional continuum (in the form of a cube say) we have the equation

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Here Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev We look for solutions in the form

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

This requires Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

From the boundary condition that Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev we get 

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

where n1; n2, n3 are nonzero positive integers.

We then get Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Each triplet (n1; n2, n3) determines a possible mode and the number of such modes whose frequency ≤ω is the volume of the all positive octant of a sphere of radius Irodov Solutions: Molecules and Crystals- 1 Notes | EduRevConsidering also the fact that the subsidiary condition div  Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev  implies two independent values of Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev for each choice of the wave vector (k1, k2, k3)

we find

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

 

Q.185. Assuming the propagation velocities of longitudinal and transverse vibrations to be the same and equal to v, find the Debye temperature
 (a) for a unidimensional crystal, i.e. a chain of identical atoms, incorporating no  atoms per unit length;
 (b) for a two-dimensional crystal, i.e. a plane square grid consisting of identical atoms, containing no  atoms per unit area;
 (c) for a simple cubic lattice consisting of identical atoms, containing no  atoms per unit volume. 

Ans. To determine the Debye temperature we cut off the high frequency modes in such a way as to get the total number of modes correctly.

(a) In a linear crystal with nol atoms, the number of modes of transverse vibrations in any given plane cannot exceed nol.  Then

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

The cut off frequency ω0 is related to the Debye temperature Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev by

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(b) In a square lattice, the number of modes of transverse oscillations cannot exceed noS.
Thus

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

or Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

(c) In a cubic crystal, the maximum number of transverse waves must be 2n0 V (two for each atom). Thus

Irodov Solutions: Molecules and Crystals- 1 Notes | EduRev

ThusIrodov Solutions: Molecules and Crystals- 1 Notes | EduRev

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