Irodov Solutions: Scattering of Particles: Rutherford-Bohr Atom - 3 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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 λ₁,λ₂ then the total energy of the excited start must be 

But  where we are ignoring reduced mass effects.

Then 

Substituting the values we get n² = 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 hydrogen-like 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 = = 5.18.10¹⁵  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
λ₁ = 0.468μm .

Q.37. Find the binding energy of an electron in the ground state of hydrogen-like 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 E₀  = 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 10⁶ m/s

Q.40. At what minimum kinetic energy must a hydrogen atom move for its inelastic head-on 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 v₁ and v₂ 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 v₂

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 Bohr-Sommerfeld 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 = ax²/2, 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 H-atom 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 λ₁ is the wavelength of the first line of Lyman series without considering nuclear motion).

Substitution gives (see 6.21 for using λ₁ = 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 λ₁ (Hydrogen) =121nm).

(b) In the positronium

Thus separation between the particles is the ground state

E_b(positronium) 

λ₁ (positronium) =  λ₂ (Hydrogen) = 0.243 nm