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# Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 1 Notes | EduRev

## JEE : Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 1 Notes | EduRev

The document Irodov Solutions: Scattering of Particles. Rutherford-Bohr Atom- 1 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.1. Employing Thomson's model, calculate the radius of a hydrogen atom and the wavelength of emitted light if the ionization energy of the atom is known to be equal to E = 13.6 eV.

Ans. The Thomson model consists of a uniformly charged nucleus in which the electrons are at rest at certain equilibrium points (the plum in the pudding model). For the hydrogen nucleus the charge on the nudeus is +e while the chaige on the electron is -e. The electron by symmetry must be at the centre of the nuclear charge where the potential (from problem (3.38a)) is

where R is the radius of the nucleur charge distribution. The potential energy of the electron is  and since the electron is at rest, this is also the total eneigy. To ionize such an electron will require an energy of

From this we find

In Gaussian system the factor  is missing.

Putting the values we get R= 0.159 nm .

Light is emitted when the electron vibrates. If we displace the electron slightly inside the nucleus by giving it a push r in some radial direction and an eneigy δE of oscillation then since the potential at a distance r in the nucleus is

the total eneigy of the nucleus becomes

or

This is the eneigy of a harmonic oscillator whose frequency is

The vibrating electron emits radiation of frequency ω whose wavelength is

In Gaussian units the factor    is missing.

Putting the values we get

Q.2. An alpha particle with kinetic energy 0.27 MeV is deflected through an angle of 60° by a golden foil. Find the corresponding value of the aiming parameter.

Ans. Equation (6.1a) of the book reads in MKS units

Thus

For α particle

(In Gaussian units there is no factor

Substituting we get          b = 0.731 pm .

Q.3. To what minimum distance will an alpha particle with kinetic energy T = 0.40 MeV approach in the case of a head-on collision to (a) a stationary Pb nucleus; (b) a stationary free Liz nucleus?

Ans.  (a) In the Pb case we shall ignore the recoil of the nucleus both because Pb is quite heavy  as well as because Pb in not free. Then for a head on collision, at the distance of closest approach, the K.E. of the α - particle must become zero (because α - particle will turn back at this point). Then

(No  in Gaussian units.). Thus putting the values

rmin = 0.591 pm .

(b) Here we have to take account of the fact that part of the energy is spent in the recoil of Li nucleus. Suppose x1 coordinate of the α - particle from some arbitrary point on the line joining it to the Li nucleus, x= coordinate of the Lnucleus with respect to the same point Then we have the energy momentum equations

We complete the square on the right hand side and rewrite the above equation as

For the least distance of apporach, the second term on the right must be greatest which implies that the first term must vanish.

Thus

Using   and other values we get

(In Gaussian units the factor  is absent).

Q.4. An alpha particle with kinetic energy T = 0.50 MeV is deflected through an angle of θ = 90° by the Coulomb field of a stationary Hg nucleus. Find:

• All the formulas in this Part are given in the Gaussian system of units.
(a) the least curvature radius of its trajectory;
(b) the minimum approach distance between the particle and the nucleus.

Ans. We shall ignore the recoil of Hg nucleus.

(a) Let A be the point of closest approach to the centre  At A the motion is instantaneously circular because the radial velocity vanishes. Then if v0 is the speed of the particle at A, the following equations hold

(2)

(3)

(This is Newton’s law. Here is the radius of curvature of the path at A and p is minimum at A by symmetry. ρ Finally we have Eqn. (6.1 a) in the form

(4)

From (2) and (3)

or

with

(b) From (2) and (4) we write

Substituting in (1)

Solving for v0 we get

Then

Q.5. A proton with kinetic energy T and aiming parameter b was deflected by the Coulomb field of a stationary Au nucleus. Find the momentum imparted to the given nucleus as a result of scattering.

Ans. By momentum conservation

Thus the momentum transfered td the gold nucleus is clearly

A lthough the momentum transfered to the Au nucleus is  not small, the energy associated with this recoil is quite small and its effect back on the motion of the proton can be neglected to a first approximation. Then

Here  is the unit vector in the direction of the incident proton and  is normal to it on the side on which it is scattered. Thus

Or       using  tan for the proton we get

Q.6. A proton with kinetic energy T = 10 MeV flies past a stationary free electron at a distance b = 10 pm. Find the energy acquired by the electron, assuming the proton's trajectory to be rectilinear and the electron to be practically motionless as the proton flies by.

Ans.  The proton moving by the electron first accelerates and then decele rates and it not easy to calculate the energy lost by the proton so energy conservation does not do the trick. Rather we must directly calculate the momentum acquire by the electron. By symmetry that momentum is along  and its magnitude is

where  is the component along OA of the force on electron. Thus

Evaluate the integral by substituting

x - 5 tan 8

Then

Then

In Gaussian units there is no factor . Substituting the values we get

Q.7. A particle with kinetic energy T is deflected by a spherical potential well of radius R and depth Uo, i.e. by the field in which the potential energy of the particle takes the form

where r is the distance from the centre of the well. Find the relationship between the aiming parameter b of the particle and the angle 0 through which it deflects from the initial motion direction.

Ans.  See the diagram on the next page. In the region where potential is nonzero, the kinetic eneigy of the particle is, by eneigy conservation, ____________ T + Uand the mom entum of the particle has the m agnitude . . On the boundary the force is radial, so the tangential component of the momentum does not change :

We also have

Therefore

or

or

or

or

Hence

Finally, the impact parameter is

Q.8. A stationary ball of radius R is irradiated by a parallel stream of particles whose radius is r. Assuming the collision of a particle and the ball to be elastic, find:
(a) the deflection angle θ of a particle as a function of its aiming parameter b;
(b) the fraction of particles which after a collision with the ball are scattered into the angular interval between θ and θ + dθ;
(c) the probability of a particle to be deflected, after a collision with the ball, into the front hemispher

Ans.  It is implied that the ball is too heavy to recoil.

(a) The trajectory of the particle is symmetrical about the radius vector through the point of impact. It is clear from the diagram that

A lso

(b) With b defined above, the fraction of particles scattered between θ and θ + dθ (or the probability of the same) is

(c) This is

Q.9. A narrow beam of alpha particles with kinetic energy 1.0 MeV falls normally on a platinum foil 1.0 µm thick. The scattered particles are observed at an angle of 60° to the incident beam direction by means of a counter with a circular inlet area 1.0 cm2  located at the distance 10 cm from the scattering section of the foil. What fraction of scattered alpha particles reaches the counter inlet?

Ans. From the formula (6.1 b) of the book

We have put  here. Also n = no. of Pt nuclei in the foil per unit area

Using the values

we get

Also

Substituting we get

Q.10. A narrow beam of alpha particles with kinetic energy T = = 0.50 MeV and intensity I = 5.0.105  particles per second falls normally on a golden foil. Find the thickness of the foil if at a distance r = 15 cm from a scattering section of that foil the flux density of scattered particles at the angle θ = 60° to the incident beam is equal to J = 40 particles/(cm2•s).

Ans. a scattered flux density of J (perticles per unit area per second) equals  particles scattered per unit time per steradian in the given direction. Let n = concentration of the gold nuclei in the foil. Then

and the number of Au nuclei per unit area of the foil is nd where d = thickness of the foil Then from Eqn. (6.1 b) (note that n → nd here)

Here I  is the number of α - particles falling on the foil per second

Hence

using Z = 79 , AAu = 197, ρ = 19-3 x 103 kg/m3 NA = 6.023 x 1026 / kilo mole and other data from the problem we get

d = 1.47 pm

Q.11. A narrow beam of alpha particles falls normally on a silver foil behind which a counter is set to register the scattered particles. On substitution of platinum foil of the same mass thickness for the silver foil, the number of alpha particles registered per unit time increased η = 1.52 times. Find the atomic number of platinum, assuming the atomic number of silver and the atomic masses of both platinum and silver to be known.

Ans. From the formula (6.1 b) of the book, wc find

But since the foils have the same mass thickness ( = pd ) , we have

see the problem (6.9). Hence

Substituting    we get

Q.12. A narrow beam of alpha particles with kinetic energy T = = 0.50 MeV falls normally on a golden foil whose mass thickness is pd = 1.5 mg/cm2. The beam intensity is I0  = 5.0.10 particles per second. Find the number of alpha particles scattered by the foil during a time interval ζ = 30 min into the angular interval:
(a) 59-61°;
(b) over θ = 60°.

Ans. (a) From Eqn. (6.1 b) we get

(we have used

From the data

Also ZAu = 79 , AAm = 197 . Putting the values we get

dN = 1.63 x 106

(b) This number is

The integral is

Thus

where n is the concentration of nuclei in the foil.

Substitution gives

Q.13. A narrow beam of protons with velocity v = 6.106  m/s falls normally on a silver foil of thickness d = 1.0 p,m. Find the probability of the protons to be scattered into the rear hemisphere (θ > 90°).

Ans. The requisite probability can be written easily by analogy with (b) of the previous problem. It is

The integral is unity. Thus

Substitution gives using

Q.14. A narrow beam of alpha particles with kinetic energy T = = 600 keV falls normally on a golden foil incorporating n = 1.1.1019  nuclei/cm2. Find the fraction of alpha particles scattered through the angles θ < θ = 20°.

Ans. Because of the   dependence of the scattering, the number of particles (or fraction) scattered through θ < θO cannot be calculated directly. But we can write this fraction as

where Q (θO) is the fraction of particles scattered through . This fraction has been calculated before and is (see the results of 6.12 (b))

where n here is number of nuclei/cm2 . Using the data we get

Q = 0.4

Thus

Q.15. A narrow beam of protons with kinetic energy T = 1.4 MeV falls normally on a brass foil whose mass thickness pd = 1.5 mg/cm2. The weight ratio of copper and zinc in the foil is equal to 7 : 3 respectively. Find the fraction of the protons scattered through the angles exceeding θ0  = 30°.

Ans. The relevant fraction can be immediately written down (see 6.12 (b)) (Note that the projectiles are protons)

Here n1 (n2) is the number of Zn(Cu) nuclei per cm2 of the foil and Z1(Z2) is the atomic number of Zn(Cu). Now

Here M i , M 2 are the mass numbers o f Zn and Cu.

Then, substituting the values Z1 = 30, Z= 29, M= 65.4, M2 = 63.5, we get

Q.16. Find the effective cross section of a uranium nucleus corresponding to the scattering of alpha particles with kinetic energy T = 1.5 MeV through the angles exceeding θ0  = 60°.

Ans. From the Rutherford scattering formula

or

Then integrating from  we get the required cross section

For U nucleus Z - 92 and we get on putting the values

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