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 = x_{o} (1. — e ^{hω/kT}), where x_{o} is the absorption coefficient for T_{0}.
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
Here N_{1 }= N_{o}. of atoms in lower level
N_{2} = N_{o} of atoms in the upper level per unit volume.
B_{12} , B_{21} are E in stein coefficients and
I_{c} = energy density in the beam ,
c = velocity of light
A factor arises because each transition result in a loss or gain of energy
Hence
But g_{1}B_{12} = g_{2}B_{21} so
By Boltzman factor
When we can put N_{1}= N_{0} the total number of atoms per unit volume.
Then
where is 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 which is transmitted to the photon it emits as natural broadening. Then
The Doppler broadening on the other hand arises from the thermal motion of radiating atoms.
The effect is nonrelativistic and the maximum broadening can be written as
Thus
Substitution gives using
Note: Our formula is an order of magnitude estimate.
Q.133. Find the wavelength of the K_{α } line in copper (Z = 29) if the wavelength of the K_{α } line in iron (Z = 26) is known to be equal to 193 pm.
Ans. From Moseley’s law
or
Thus
Substitution gives
Q.134. Proceeding from Moseley's law find:
(a) the wavelength of the K_{α } line in aluminium and cobalt:
(b) the difference in binding energies of K and L electrons in vanadium.
Ans. (a) From Moseley’s law
or
We shall take σ = 1. For Aluminium
and for cobalt (= 27)
(b) This difference is nearly equal to the energy of the K_{α} line which by Moseley’s law is equal to (= 23 for vanadium)
Q.135. How many elements are there in a row between those whose wavelengths of K_{α }lines are equal to 250 and 179 pm?
Ans. We calculate the values corresponding to the given wavelengths using Moseley’s law. See problem (134).
Substitution gives that
= 23 corresponding to λ = 250 pm
and = 27 corresponding to λ = 179 pm
There are thus three elements in a row between those whose wavelengths of K_{α} lines are equal to 250 pm and 179 pm.
Q.136. Find the voltage applied to an Xray tube with nickel anticathode if the wavelength difference between the K_{α } line and the shortwave cutoff of the continuous Xray spectrum is equal to 84 pm.
Ans. From Moseley’s law
where = 28 for N_{i} . Substitution gives
Now the short wave cut of off the continuous spectrum must be more energetic (smaller wavelength) otherwise K_{α} lines will not emerge. Then since we get
λ = 82.5 pm
This corresponds to a voltage of
Substitution gives V = 15.0 kV
Q.137. At a certain voltage applied to an Xray tube with aluminium anticathode the shortwave cutoff wavelength of the continuous Xray 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
λ_{o} = 0.50nm
the voltage applied m ust be
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 Xray tube increased from V_{1} =10kV to V_{2} = 20 kV, the wavelength interval between the K_{α} line and the shortwave cutoff of the continuous Xray 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 λ_{o} = wavelength of the characteristic Xray line. Then using the formula for short wavelength limit of continuous radiation
Hence
Using also Moseley’s law, we get
Q.139. What metal has in its absorption spectrum the difference between the frequencies of Xray K and L absorption edges equal to Δω = 6.85.10^{18} 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 K_{α} line (see the diagram below). Thus by Moseley’s formule
or
The metal is titanium,
Q.140. Calculate the binding energy of a K electron in vanadium whose L absorption edge has the wavelength λ_{L} = 2.4 nm.
Ans. From the diagram above w e see th at the binding energy E_{b} of a K electron is the sum o f the energy of a K_{α} line and the energy corresponding to the L edge of absorption spectrum
For vanadium = 23 and the energy of K_{α} line of vanadium has been calculated in problem 134 (b). Using
we get E_{b} = 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 shortwave cutoff is Δλ = 26 pm.
Ans. By Moseley’s law
where  E_{K} is the energy of the K electron and  E_{L }of the L electron. Also the energy of the line corresponding to the short wave cut off of the K series is
Hence
Substitution gives for titanium ( = 22)
ω = 6.85 x 10^{18} s^{1}
and hence E_{L }= 0.47keV
Q.142. Find the kinetic energy and the velocity of the photoelectrons liberated by K_{α }radiation of zinc from the K shell of iron whose K band absorption edge wavelength is λ_{K} = 174 pm.
Ans. The energy of the K a radiation of _{n} is
where E = atomic number of zinc = 30. Th e binding energy of the K electrons in iron is obtained from the wavelength of K absorption edge as. Hence by Einstein equation
Substitution gives
T = 1.463 k eV
This corresponds to a velocity of the photo electrons of
v = 2.27x10^{6}m/s
Q.143. Calculate the Lande g factor for atoms (a) in S states; (b) in singlet states.
Ans. From the Lande formula
(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
Q.144. Calculate the Lande g factor for the following terms: (a) ^{6}F_{1/2}; (b) ^{4}D_{1/2}; (c) ^{5}F_{2}; (d) ^{5}P_{1}; (e) ^{3}P_{o}.
Ans. (a) Here
(b) Here
(c) ^{5}F_{2} Here S = 2 , L = 3 , J = 2
(d) ^{5}P_{1} Here S = 2 , L = 1 , J = 1
(e) ^{3}P_{0}. 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 ^{1}F state; (b) in ^{2}D_{3/2} 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
Hence
(b) For the
Hence
(c) We have
or
or
Hence
Q.146. Determine the spin angular momentum of an atom in the state D_{2} 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
where mj is the projection of
Maximum value of the m_{j }is J. Thus
gJ = 4
Since J = 2 , we get g = 2. Now
Hence S (S + 1) = 12 or S = 3
Thus
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.
Thus
or
Hence J = 3
Then
and
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
Then
Hence
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