Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

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

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Q.131. A beam of light of frequency ω, equal to the resonant frequency of transition of atoms of gas, passes through that gas heated to temperature T. In this case hω >> kT. Taking into account induced radiation, demonstrate that the absorption coefficient of the gas x varies as x = xo  (1. — e -hω/kT), where xo  is the absorption coefficient for T0

Ans.  Because of the resonant nature of the processes we can ignore nonresonant processes. We also ignore spontaneous emission since it does not contribute to the absorption coefficient and is a small term if the beam is intense enough.
Suppose I is the intensity of the beam at some point. The decrease in the value of this intensity on passing through the layer of the substance of thickness d x is equal to

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Here N= No. of atoms in lower level
N2 = No of atoms in the upper level per unit volume.
B12 , B21 are E in stein coefficients and
Ic = energy density in the beam ,
c = velocity of light

A factor Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev arises because each transition result in a loss or gain of energy Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

But     g1B12 = g2B21 so

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

By Boltzman factor Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

When Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev we can put N1= N0 the total number of atoms per unit volume. 

Then Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevis the absorption coefficient for T → 0.

 

Q.132. The wavelength of a resonant mercury line is λ = = 253.65 nm. The mean lifetime of mercury atoms in the state of resonance excitation is ζ = 0.15 μs. Evaluate the ratio of the Doppler line broadening to the natural linewidth at a gas temperature T = 300 K.

Ans. A short lived state of mean life T has an uncertainty in energy of  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev which is transmitted to the photon it emits as natural broadening. Then

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

The Doppler broadening on the other hand arises from the thermal motion of radiating atoms.
The effect is non-relativistic and the maximum broadening can be written as

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Thus Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Substitution gives using  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Note: Our formula is an order of magnitude estimate.

 

Q.133. Find the wavelength of the Kα  line in copper (Z = 29) if the wavelength of the Kα  line in iron (Z = 26) is known to be equal to 193 pm. 

Ans. From Moseley’s law

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Thus  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Substitution gives

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.134. Proceeding from Moseley's law find:
 (a) the wavelength of the Kα   line in aluminium and cobalt: 
 (b) the difference in binding energies of K and L electrons in vanadium. 

Ans. (a) From Moseley’s law

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or   Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

We shall take σ = 1. For Aluminium  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

and for cobalt  (Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev= 27)

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(b) This difference is nearly equal to the energy of the Kα line which by Moseley’s law is equal to (Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev= 23 for vanadium)

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.135. How many elements are there in a row between those whose wavelengths of Kα lines are equal to 250 and 179 pm?

Ans. We calculate the Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev values corresponding to the given wavelengths using Moseley’s law. See problem (134).
Substitution gives that

        Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev= 23 corresponding to λ = 250 pm 

and   Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev= 27 corresponding to λ = 179 pm

There are thus three elements in a row between those whose wavelengths of Kα lines are equal to 250 pm and 179 pm.

 

Q.136. Find the voltage applied to an X-ray tube with nickel anticathode if the wavelength difference between the Kα  line and the short-wave cut-off of the continuous X-ray spectrum is equal to 84 pm. 

Ans. From Moseley’s law

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

where Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev = 28 for Ni . Substitution gives

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Now the short wave cut of off the continuous spectrum must be more energetic (smaller wavelength) otherwise Kα lines will not emerge. Then since Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev we get

λ = 82.5 pm

This corresponds to a voltage of

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Substitution gives V = 15.0 kV

 

Q.137. At a certain voltage applied to an X-ray tube with aluminium anticathode the short-wave cut-off wavelength of the continuous X-ray spectrum is equal to 0.50 nm. Will the K series of the characteristic spectrum whose excitation potential is equal to 1.56 kV be also observed in this case? 

Ans. Since the short wavelength cut off of the continuous spectrum is

λo = 0.50nm

the voltage applied m ust be Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

since this is greater than the excitation potential of the K series of the characteristic spectrum (which is only 1.56 k V ) the latter will be observed.

 

Q.138. When the voltage applied to an X-ray tube increased from V1  =10kV to V2 = 20 kV, the wavelength interval between the Kα  line and the short-wave cut-off of the continuous X-ray spectrum increases by a factor n = 3.0. Find the atomic number of the element of which the tube's anticathode is made. 

Ans. Suppose λo = wavelength of the characteristic X-ray line. Then using the formula for short wavelength limit of continuous radiation

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Using also Moseley’s law, we get

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.139. What metal has in its absorption spectrum the difference between the frequencies of X-ray K and L absorption edges equal to Δω = 6.85.1018 s-1 

Ans. The difference in frequencies of the K and L absorption edges is equal, according to the Bohr picture, to the frequency of the Kα line (see the diagram below). Thus by Moseley’s formule

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

The metal is titanium,

 

Q.140. Calculate the binding energy of a K electron in vanadium whose L absorption edge has the wavelength λL = 2.4 nm. 

Ans. From the diagram above w e see th at the binding energy Eb of a K electron is the sum o f the energy of a Kα line and the energy corresponding to the L edge of absorption spectrum

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

For vanadium Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev = 23 and the energy of Kα line of vanadium has been calculated in problem 134 (b). Using

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

we get                   Eb = 5.46 k eV

 

Q.141. Find the binding energy of an L electron in titanium if the wavelength difference between the first line of the K series and its short-wave cut-off is Δλ = 26 pm. 

Ans. By Moseley’s law

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

where - EK is the energy of the K electron and - Eof the L electron. Also the energy of the line corresponding to the short wave cut off of the K series is

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Substitution gives for titanium ( Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev= 22)

ω = 6.85 x 1018 s-1

and hence E= 0.47keV

 

Q.142. Find the kinetic energy and the velocity of the photoelectrons liberated by Kα radiation of zinc from the K shell of iron whose K band absorption edge wavelength is λK = 174 pm. 

Ans. The energy of the K a radiation of Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevn is

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

where E = atomic number of Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev zinc = 30. Th e binding energy of the K electrons in iron is obtained from the wavelength of K absorption edge asIrodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev.  Hence by Einstein equation

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Substitution gives

T = 1.463 k eV

This corresponds to a velocity of the photo electrons of

v = 2.27x106m/s

 

Q.143. Calculate the Lande g factor for atoms (a) in S states; (b) in singlet states. 

Ans. From the Lande formula

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(a) For S states L = 0. This implies J = S. Then, if S = 0

g = 2

(For singlet states g is not defined if L = 0 )

(b) For singlet states, J = L

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.144. Calculate the Lande g factor for the following terms: (a) 6F1/2; (b) 4D1/2; (c) 5F2; (d) 5P1; (e) 3Po

Ans. (a)  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev Here   Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(b) Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevHere  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(c) 5F2 Here S = 2 , L = 3 , J = 2

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(d) 5P1 Here S = 2 , L = 1 , J = 1

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(e) 3P0. For states with J = 0, L = S the g gactor is indeterminate.

 

Q.145. Calculate the magnetic moment of an atom (in Bohr magnetons) (a) in 1F state; (b) in 2D3/2  state; (c) in the state in which S = 1, L = 2, and Lande factor g = 4/3. 

Ans. (a) For the lF state S = 0, L = 3 = J

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

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

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

(c) We have Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.146. Determine the spin angular momentum of an atom in the state D2 if the maximum value of the magnetic moment projection in that state is equal to four Bohr magnetons. 

Ans.  The expression for the projection of the magnetic moment is

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

where mj is the projection of Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Maximum value of the mis J. Thus

gJ = 4

Since J = 2 , we get g = 2. Now

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence      S (S + 1) = 12 or S = 3

Thus Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.147. An atom in the state with quantum numbers L = 2, S = 1. is located in a weak magnetic field. Find its magnetic moment if the least possible angle between the angular momentum and the field direction is known to be equal to 30°. 

Ans. The angle between the angular momentum vector and the field direction is the least when the angular mommentum projection is maximum i.e.  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Thus  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

or Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Hence         J = 3

Then Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRevIrodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

and  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

 

Q.148. A valence electron in a sodium atom is in the state with principal quantum number n = 3, with the total angular momentum being the greatest possible. What is its magnetic moment in that state? 

Ans. For a state with n = 3 , l = 2. Thus the state with maximum angular momentum is

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Then Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

HenceIrodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev  Irodov Solutions: Properties of Atoms. Spectra- 3 Notes | EduRev

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