Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Physics For JEE

: Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

The document Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev is a part of the Course Physics For JEE.

Q.17. The effective cross section of a gold nucleus corresponding to the scattering of monoenergetic alpha particles within the angular interval from 90° to 180° is equal to Δσ = 0.50 kb. Find:
 (a) the energy of alpha particles;
 (b) the differential cross section of scattering dσ/dΩ (kb/sr) corresponding to the angle θ = 60°. 

Ans. (a) From the previous formula

 Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

or   Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Substituting the values with Z = 79 we get (θO = 90°)

T = 0.903 MeV

(b) The differential scattering cross section is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus from the given data

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

So

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.18. In accordance with classical electrodynamics an electron moving with acceleration ω loses its energy due to radiation as 

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where e is the electron charge, c is the velocity of light. Estimate the time during which the energy of an electron performing almost harmonic oscillations with frequency ω = 5.1015  s-1  will decrease η = 10 times.

Ans. The formula in MKS units is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

For an electron performing (linear) harmonic vibrations Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev is in some definite directions with

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

If the radiation loss is small (i.e. if ω is not too large), then the motion of the electron is always close to simple harmonic with slowly decreasing amplitude. Then we can write

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and average the above equation ignoring the variation of a in any cycle. Thus we get the equation, on using 

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

since  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev for a harmonic oscillator.

This equation integrates to

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

It is then seen that energy decreases η times in

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.19. Making use of the formula of the foregoing problem, estimate the time during which an electron moving in a hydrogen atom along a circular orbit of radius r = 50 pm would have fallen onto the nucleus. For the sake of simplicity assume the vector ω to be permanently directed toward the centre of the atom. 

Ans. Moving around the nucleus, the electron radiates and its energy decreases. This means that the electron gets nearer the nucleus. By the statement of the problem we can assume that the electron is always moving in a circular orbit and the radial acceleration by Newton’s law is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

directed inwards. Thus

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

On the other hand in a circular orbit

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

so

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

or

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Integrating

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and the radius fells to zero in

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.20. Demonstrate that the frequency ω of a photon emerging when an electron jumps between neighbouring circular orbits of a hydrogen-like ion satisfies the inequality ωn > ω > ωn +1, where ωn  and ωn +1 are the frequencies of revolution of that electron around the nucleus along the circular orbits. Make sure that as n   → Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev the frequency of the photon ω → ωn.

Ans. In a circular orbit we have the following formula

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Then Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

The energy E is Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and the circular frequency of this orbit is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

On the other hand the frequency co of the light emitted when the electron makes a transition n + 1 →  n is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus the inequality

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

will result if

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Or multiplying by n2(n + 1)2 we have to prove

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

This can be written as

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

This is obvious because  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

For laige n

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

soIrodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.21. A particle of mass m moves along a circular orbit in a centrosymmetrical potential field U (r) = kr212. Using the Bohr quantization condition, find the permissible orbital radii and energy levels of that particle. 

Ans. We have the following equation (we ignore reduced mass effects)


Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

so  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 and Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and   Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

The energy levels are  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.22. Calculate for a hydrogen atom and a He + ion:
 (a) the radius of the first Bohr orbit and the velocity of an electron moving along it;
 (b) the kinetic energy and the binding energy of an electron in the ground state;
 (c) the ionization potential, the first excitation potential and the wavelength of the resonance line (n' = 2  →  n = 1). 

Ans. The basic equations have been derived in the problem (6.20). We rewrite them here and determine the the required values.

(a) Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev Thus Z = 1 for H , Z = 2 for He+

Thus                     r1 = 52.8 pm, for H atom
                               r= 26-.4 pm, for He+ ion

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

v= 2.191 x 106 m/s for H atom

= 4.382 x 106 m/s for He+ ion

(b)   Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

T = 13.65 eV for H atom T = 54.6 eV for He+ ion

In both cases E h = T because Eb = - E and E = - T (Recall that for coulomb force V = - 2 T )

(c) The ionization potential φ is given by

= E

so φ = 13.65 volts for H atom

φ= 54.6 volts for He+ ion 13.65

The energy levels are

The energy levels are Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev  for H atom

and Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev = 10.23 volts for H atom

= 4 x 10.23 = 40.9 volts for He+ ion 

The wavelength of the resonance line

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev  is given by

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev 10.23 eV for H atom

so   λ = 121.2 nm for H atom

For He+ ion Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.23. Calculate the angular frequency of an electron occupying the second Bohr orbit of He+ ion. 

Ans. This has been calculated before in problem (6.20). It is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.24. For hydrogen-like systems find the magnetic moment tt n corresponding to the motion of an electron along the n-th orbit and the ratio of the magnetic and mechanical moments µn /Mn. Calculate the magnetic moment of an electron occupying the first Bohr orbit.

Ans. An electron moving in a circle with a time period T constitutes a current

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and forms a current loop of area πr2. This is equivalent to magnetic moment,

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

on using v = 2πr/T. Thus

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

for the nth orbit. (In Gaussian units

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

We see that

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev is the angular momentum 

Thus  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

(In CGS units Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.25. Calculate the magnetic field induction at the centre of a hydrogen atom caused by an electron moving along the first Bohr orbit. 

Ans. The revolving electron is equivalent to a circular current

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

The magnetic induction

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Substitution gives B = 12.56 T at the centre. (In Gaussian units

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.26. Calculate and draw on the wavelength scale the spectral intervals in which the Lyman, Balmer, and Paschen series for atomic hydrogen are confined. Show the visible portion of the spectrum. 

Ans. From the general formula for the transition n2 → n1

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where EH ≥ 13.65eV. Then

(1) Lyman, n= 1, n2 = 2,3. Thus

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

This conesponds to Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

and Lyman lines have λ ≤ 0.121μm with the series limit at .0909μm

(2) Balmer : n2 = 2 , n3 = 3, 4 ,

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

This corresponds to
λ = 0.65 μm

and Balmer series has λ ≤ 0.65μm with the series limit at X = 0.363μm

(3) Paschen : n2 = 3 , n= 4, 5,...

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

This corresponds to λ = 1.869μm

with the series limit at λ = 0.818μm

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.27. To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series: 486.1 and 410.2 nm? What is the wavelength of that line? 

Ans. The Balmer line of wavelength 486-1 nm is due to the transition 4 → 2 while the Balmer line of wavelingth 4→2 nm is due to the transition 6→2. The line whose wave number corresponds to the difference in wave numbers of these two lines is due to the transition 6 → 4. That line belongs to the Brackett series. The wavelength of this line is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.28. For the case of atomic hydrogen find:
 (a) the wavelengths of the first three lines of the Balmer series;
 (b) the minimum resolving power λ/δλ of a spectral instrument capable of resolving the first 20 lines of the Balmer series. 

Ans. The energies are

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

They correspond to wavelengths 654.2 n m , 484.6 n m and 433 nm

The nth line of the Balmer series has the energy

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

For n = 19, we get the wavelength 366.7450 nm

For n = 20 we get the wavelength 366.4470 nm

To resolve these lines we require a resolving power of

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

 

Q.29. Radiation of atomic hydrogen falls normally on a diffraction grating of width l = 6.6 mm. The 50th line of the Balmer series in the observed spectrum is close to resolution at a diffraction angle θ (in accordance with Rayleigh's criterion). Find that angle. 

Ans. For the Balmer series

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev Thus

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

or Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

or Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

On the other hand for just resolution in a diffraction grating

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

HenceIrodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Substitution gives θ = 59.4°.

 

Q.30. What element has a hydrogen-like spectrum whose lines have wavelengths four times shorter than those of atomic hydrogen? 

Ans. If all wavelengths are four times shorter but otherwise similar to the hydrogen atom spectrum then the energy levels of the given atom must be four times greater.

This means Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev  

compared to  Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev for hydrogen atom. Therefore the spectrum is that o f He+ ion (Z = 2).

 

Q.31. How many spectral lines are emitted by atomic hydrogen excited to the n-th energy level?

Ans. Because of cascading all possible transitions are seen. Thus we look for the number of ways in which we can select upper and lower levels. The number of ways we can do this is

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

where the factor Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev takes account of the fact that the photon emission always arises from upper lower transition.

 

Q.32. What lines of atomic hydrogen absorption spectrum fall within the wavelength range from 94.5 to 130.0 nm? 

Ans. These are the Lyman lines

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 2 Notes | EduRev

Thus at the level of accuracy of our calculation, there are four lines 121.1nm , 102.2nm , 96.9nm and 94.64nm .

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