Q.33. Find the quantum number n corresponding to the excited state of He^{+} ion if on transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm.
Ans. If the wavelengths are λ_{1},λ_{2} then the total energy of the excited start must be
But where we are ignoring reduced mass effects.
Then
Substituting the values we get n^{2} = 23
which we take to mean n = 5. (The result is sensitive to the values of the various quantities and small differences get multiplied because difference of two large quantities is involved :
Q.34. Calculate the Rydberg constant R if He^{+} ions are known to have the wavelength difference between the first (of the longest wavelength) lines of the Balmer and Lyman series equal to Δλ = = 133.7 nm.
Ans. For the longest wavelength (first) line of the Balmer series we have on using the generalized Balmer formula
the result
Then
so
Q.35. What hydrogenlike ion has the wavelength difference between the first lines of the Balmer and Lyman series equal to 59.3 nm?
Ans. From the formula of the previous problem
or
Substitution of Δλ = 59.3nm and and the previous problem gives Z = 3 This identifies the ion as Li^{++}
Q.36. Find the wavelength of the first line of the He ion spectral series whose interval between the extreme lines is Δω = = 5.18.10^{15 } s^{1},
Ans. We start from the generalized Balmer formula
Here m = n + 1,n + 2, ....
The interval between extreme lines of this series (series n) is
Hence
Then the angular frequency of the first line of this series (series n) is
Then the wavelength will be
Substitution (with the value of R from problem 6.34 which is also the correct value determined directly) gives
λ_{1} = 0.468μm .
Q.37. Find the binding energy of an electron in the ground state of hydrogenlike ions in whose spectrum the third line of the Balmer series is equal to 108.5 nm.
Ans. For the third line of of Balmer series
Hence
or
Substitution gives Z = 2. Hence the binding energy of the electron in the ground state of this ion is
E_{b }= 4E_{H} = 4 x 13.65 = 54.6 eV
The ion is He^{+}
Q.38. The binding energy of an electron in the ground state of He^{+} atom is equal to E_{0 } = 24.6 eV. Find the energy required to remove both electrons from the atom.
Ans. To remove one electron requires 24.6 eV.
The ion that is left is He^{+} which in its ground start has a binding energy of
The complete binding energy of both electrons is then
Substitution gives E = 79.1 eV
Q.39. Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength λ = 18.0 nm from stationary He ions in the ground state.
Ans. By conservation of energy
where is the binding energy of the electron in the ground state of He^{+}. (Recoil of He^{++} nucleus is neglected). Then
Substitution gives
v = 2.25 x 10^{6 }m/s
Q.40. At what minimum kinetic energy must a hydrogen atom move for its inelastic headon collision with another, stationary, hydrogen atom to make one of them capable of emitting a photon? Both atoms are supposed to be in the ground state prior to the collision.
Ans. Photon can be emitted in H  H collision only if one o f the H is excited to an n = 2 state which then dexcites to n = 1 state by emitting a photon. Let v_{1} and v_{2} be the velocities of the two Hydrogen atoms after the collision and M their masses. Then, energy momentum conservation
(in the frame of the stationary H atom)
is the excitation energy of the n = 2 state from the ground state.
Eliminating v_{2}
or
or
For minimum T, the square on the left should vanish. Thus
Q.41. A stationary hydrogen atom emits a photon corresponding to the first line of the Lyman series. What velocity does the atom acquire?
Ans. In the rest frame of die original excited nudeus we have the equations
is the eneigy available in transition corresponding to the first Lyman line.)
Then
or
(We could have written this directly by noting that Then
Q.42. From the conditions of the foregoing problem find how much (in per cent) the energy of the emitted photon differs from the energy of the corresponding transition in a hydrogen atom.
Ans. We have
Then
Q.43. A stationary He ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. Find the velocity of the photoelectron.
Ans. We neglect recoil effebts. The energy of the first Lyman line photon emitted by He^{+} is
The velocity v of the photoelectron that this photon liberates is given by
where h R on the right is the binding eneigy of the n = 1 electron in H atom. Thus
Here m is the mass of the electron.
Q.44. Find the velocity of the excited hydrogen atoms if the first line of the Lyman series is displaced by Δλ = 0.20 nm when their radiation is observed at an angle θ = 45° to their motion direction.
Ans. Since of the first Lyman line of H atom, we need not worry about effects. Then
Hence
or
But
Hence
Substitution gives
Q.45. According to the BohrSommerfeld postulate the periodic motion of a particle in a potential field must satisfy the following quantization rule:
where q and p are generalized coordinate and momentum of the particle , n are integers. Making use of this rule, find the permitted values of energy for a particle of mass m moving
(a) in a unidimensional rectangular potential well of width l with infinitely high walls;
(b) along a circle of radius r;
(c) in a unidimensional potential field U = ax^{2}/2, where a is a positive constant;
(d) along a round orbit in a central field, where the potential energy of the particle is equal to U = —α/r (α is a positive constant).
Ans. (a) If we measure eneigy from the bottom of the well, then V(x) = 0 inside the walls. Then
the quantization condition reads
or
Hence
because we have to consider the integral form and then back to
(b) Here
or
Hence
(c) By energy conservation
so
The integral is
Thus
Hence
(b) It is requined to find the energy levels of the circular orbait for the rotential
In a circular orbit, the particle only has tangential velocity and the quantization condition read
so
The energy of the particle is
Equilibrium requires that the energy as a function of r be minimum. Thus
Hence
Q.46. Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state and for the Rydberg constant. How much (in per cent) do the binding energy and the Rydberg constant, obtained without taking into account the motion of the nucleus, differ from the more accurate corresponding values of these quantities?
Ans. The total energy of the Hatom in an arbitrary frame is
Here are the coordinates of the electron and protons.
We define
Then
or
and we get
In the frame this reduces to the energy of a particle of mass
μ is called the reduced mass.
Then
Since
these values differ by (= 0.54%) from the values obtained without considering nuclear motion (M = 1837m)
Q..47. For atoms of light and heavy hydrogen (H and D) find the difference
(a) between the binding energies of their electrons in the ground state;
(b) between the wavelengths of first lines of the Lyman series.
Ans. The difference between the binding energies is
Substitution gives .
For the first line of the Lyman series
or
Hence
(where λ_{1} is the wavelength of the first line of Lyman series without considering nuclear motion).
Substitution gives (see 6.21 for using λ_{1} = 121 nm
Q.48. Calculate the separation between the particles of a system in the ground state, the corresponding binding energy, and the wavelength of the first line of the Lyman series, if such a system is
(a) a mesonic hydrogen atom whose nucleus is a proton (in a mesonic atom an electron is replaced by a meson whose charge is the same and mass is 207 that of an electron);
(b) a positronium consisting of an electron and a positron revolving around their common centre of masses.
Ans. (a) In the mesonic system, the reduced mass of the system is related to the masses of the meson (m_{μ}) and proton (m_{p}) by
Then,
separation between the particles in the ground state
= 0.284 pm
= 2.54 keV
(on using λ_{1 }(Hydrogen) =121nm).
(b) In the positronium
Thus separation between the particles is the ground state
E_{b}(positronium)
λ_{1 }(positronium) = λ_{2} (Hydrogen) = 0.243 nm
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