I.E. Irodov Solutions on Scattering of Particles Rutherford-Bohr Atom

Scattering of Particles Rutherford-Bohr Atom is a chapter that mainly deals with the scattering of particles. The students of Class 12 must pay attention to this topic as it plays a key role in understanding the basic concepts of atoms and other particles. IE IRODOV Physics Solutions provide solutions to various problems related to atoms and molecules. This helps in better understanding and grip over the subject.

The in-depth explanation of problems with related equations and diagrams aids the students to grasp easily and the concepts are engraved better in their minds. The problems and solutions provided here help the student to clear the competitive exams easily as the solutions go in flow with the theoretical points and relate with the practical applications. The numerical are solved in a complete manner with appropriate formulae, equations, diagrams and units. A diverse variety of questions are solved to help the students to practise and score higher marks in their upcoming exams.

I.E. Irodov Solutions on Scattering of Particles Rutherford-Bohr Atom

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.

    Solution:

    1. 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 nucleus is + e while the change on the electron is electron is -e. The electron by symmetry must be at the centre of the nuclear charge where the potential is

      ϕ0=(14πεo)(3e2R)\phi _{0}=(\frac{1}{4\pi \varepsilon _{o}})(\frac{3e}{2R})

      where R is the radius of the nuclear charge distribution. The potential energy of the electron is -eϕo and since the electron is at rest, and this is also the total energy. To ionise such an energetic electron will require an energy of E=eϕo

      From this we find

      R=(14πεo)(3e22E)R=(\frac{1}{4\pi \varepsilon _{o}})(\frac{3e^{2}}{2E})

      In Gussian system the factor 14πεo\frac{1}{4\pi \varepsilon _{o}} is missing

      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 energy δE of oscillation then since the potential at a distance r in the nucleus is

      IE IRODOV Chapter 6.1 Solutions

      This is the energy of a harmonic oscillator whose frequency is

      ω=14πεoe2mR3\omega =\frac{1}{4\pi \varepsilon _{o}}\frac{e^{2}}{mR^{3}}

      The vibrating electron emits radiation of frequency ω whose wavelength is

      λ=2πcω=2πcemR3(4πεo)12\lambda =\frac{2\pi c}{\omega}=\frac{2\pi c}{e}\sqrt{mR^{3}}(4\pi \varepsilon _{o})^{\frac{1}{2}}

      In Gaussian units the factor (4πεo)12(4\pi \varepsilon _{o})^{\frac{1}{2}} is missing

      Putting the values we get λ=0.237μm.


    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.

      Solution:

      1. IE IRODOV Chapter scattering of particles solutions


      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?

        Solution:

        1. (a) In the Pb case, we shall ignore the recoil of the nucleus both because Pb is quite heavy(Apb= 208= 52 X AHe) 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

          2Ze2(4πε0)rmin=T\frac{2Ze^{2}}{(4\pi \varepsilon _{0})r_{min}}=T

          In Gaussian units the factor (4πεo)(4\pi \varepsilon _{o}) is missing. Om substituting values we get

          rmin= 0.591pm

          (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, x2 = coordinate of the Li nucleus with respect to the same point Then we have the energy-momentum equations

          IE IRODOV Chapter scattering of particles solution

          Using m1/m2= 4/7 and substituting other values we get

          |x1-x2|=034 pm.

          In Gaussian units the factor (4πεo)(4\pi \varepsilon _{o}) is missing


        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

          Solution:

          1. We shall ignore the recoil of Hg nucleus.

            (a) Let A be the point of closest approach to the centre C,AC=rmin

            At A the motion is instantaneously circular because the radial velocity vanishes. Then if vo the particle at A, the following equations hold

            IE IRODOV Solutions Scattering of Particles Rutherford-Bohr Atom Question 4

            IE IRODOV Chapter 6.1 Question 4 Solution

            With z1=2 and z2,=80 we get

            ρmin= 0.231pm

            *we used, δ=δmin is the radius of curvature of the path at A and P is minimum at A by symmetry and finally found equation (4)

            (b) From equations 2 and 4 we get

            IE IRODOV Chapter 6.1 Problem 4 Solution


          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.

            Solution:

            1. By momentum conservation

              P+Pi=P+Pf\vec{P}+\vec{P_{i}}=\vec{P}+\vec{P_{f}}

              Thus the momentum transferred to the gold nucleus is clearly

              ΔP=PfPi=PP\vec{\Delta P}= \vec{P_{f}} -\vec{P_{i}}=\vec{P}-\vec{P}

              IE IRODOV Solutions Scattering of Particles Rutherford-Bohr Atom Question 5

              Although the momentum transferred 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

              ΔP=2mT(1cosθ)i^2mTsinΘj^\vec{\Delta P}= \sqrt{2mT}(1-\cos\theta )\hat{i}\sqrt{2mT}\sin\Theta \hat{j}

              Here i^\hat{i} is the unit vector in the direction of the incident proton and j^\hat{j} is normal to it on the side on which it is scattered. Thus,

              IE IRODOV Chapter 6.1 Question 5 Solution


            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.

              Solution:

              1. The proton moving by the electron first accelerates and then decele rates and it is not easy to calculate the energy lost by the pro the componen t along OA

                IE IRODOV Chapter 6.1 Question 6 Solution

                IE IRODOV Solutions Scattering of Particles Rutherford-Bohr Atom Question 6

                Evaluate the integral by substituting

                X-5 tan 8

                IE IRODOV Chapter 6.1 Problem 6 Solution

                In Gaussian units the factor (4πεo)2(4\pi \varepsilon _{o})^{2} is missing

                Te= 3.82 eV


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

              U={0for,r>RU0for,r<RU=\left\{\begin{matrix}0 for, r>R & \\ & -U_{0} for, r<R\end{matrix}\right.

              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.

                Solution:

                1. In the region where potential is nonzero, the kinetic energy of the particle is, by energy conservation, T + U0 and the momentum of the particle has the magnitude 2m(T+U0)\sqrt{2m(T+U_{0})}. On the boundary the force is radial, so the tangential component of the momentum does not change:

                  √(2mT) sinα = √2m(T-U0) sinϕ

                  So, sinϕ = √(T/T+U0) sinα = sinα/n

                  Where n = √(1+U0/T)

                  Θ = 2(α-ϕ)

                  IE IRODOV Solutions Scattering of Particles Rutherford-Bohr Atom Question 7

                  sin(θ2)=sin(αϕ)=sinαcosϕcosαsinϕsin(\frac{\theta}{2})=sin(\alpha-\phi)= sin \alpha cos \phi-cos\alpha sin\phi

                  = sinα(cos ϕ - cosα/n)

                  or nsin(θ/2)sinα=n2sin2αcosα\frac{n sin(\theta/2)}{sin \alpha}=\sqrt{n^2-sin^2\alpha}-cos \alpha

                  squaring both sides and solving, we have

                  cot α = [n cos(θ/2)-1]/nsin(θ/2)

                  Hence, sin α = [n sin(θ/2)]/√[1+n2-2n cos(θ/2)]

                  Finally, the impact parameter is

                  b=Rsinα=nRsinΘ21+n22ncosΘ2b=R\sin\alpha =\frac{nR\sin\frac{\Theta }{2}}{\sqrt{1+n^{2}-2n\cos\frac{\Theta }{2}}}


                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 hemisphere. Θ<π2\Theta < \frac{\pi }{2}

                  Solution:

                  1. 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

                    IE IRODOV Chapter 6.1 Question 8 Solution

                    Θ=π2ϕ\Theta =\pi -2\phi

                    Also b=(R+r)sinϕ=(R+r)cosΘ2b=(R+r)\sin\phi = (R+r)\cos\frac{\Theta }{2}

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

                    dP = |2πbdb|/π(R+r)2=(1/2) sinθ dθ

                    (c) P=0π/2(1/2)sinθdθ=(1/2)10d(cosθ)=1/2P =\int_0^{\pi/2}(1/2) sin\theta d\theta = (1/2)\int_{-1}^0d(-cos\theta) = 1/2


                  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 of 1.0 cm2 located at a distance of 10 cm from the scattering section of the foil. What fraction of scattered alpha particles reaches the counter inlet?

                    Solution:

                    1. We know that

                      dNN=n(ze24πϵo)2T)2dQsin4(θ/2)\frac{dN}{N}=n(\frac{ze^2}{4 \pi \epsilon_o)2T})^2 \frac{dQ}{sin^4(\theta/2)}

                      on substituting q1 =2e, q2=ze

                      Here n= number of platinum nuclei in the foil per unit area

                      IE IRODOV Book Solutions for Chapter 6.1


                    10. A narrow beam of alpha particles with kinetic energy T = = 0.50 MeV and intensity I = 5.0.105particles 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).

                      Solution:

                      1. A scattered flux density of J (particles per unit area per second) equals J1r2=r2J\frac{J}{\frac{1}{r^{2}}}=r^{2}J particles scattered per unit time per steradian in the given direction. Let n = concentration of the gold nuclei in the foil. Then

                        n=NaρAAun=\frac{N_{a}\rho }{A_{Au}}

                        and the number of Au nuclei per unit area of the foil is nd, where d = thickness of the foil

                        Then

                        IE IRODOV Chapter 6.1 Question 10 Solution

                        Using Z=79, Aau= 197, ρ= 193X 103kg/m3, NA= 6.023 X 1026 kilo mole and other data from the problem we get

                        d=1.47pm


                      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.

                        Solution:

                        1. We know that

                          dNPt/dNAg = nPt/nAg . = η

                          The foils have same mass thickness=pd, we get

                          nPt/nAg = AAg/APt

                          Hence, ZPt = ZAg √[ηAAg/APt]

                          Put ZAg = 47, AAg = 108


                        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 = 5.0.105 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 θ0 = 60°

                          Solution:

                          1. We know that

                            IE IRODOV Chapter 6 Question 12

                            Also Z Au = 79 , AAm=197

                            On substituting these values we get

                            dN = 1.63 x 106

                            (b) The number is

                            IE IRODOV Chapter 6 Question 12 Solution


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

                            Solution:

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

                              IE IRODOV Chapter 6 Question 13


                            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 < θ0 = 20°.

                              Solution:

                              1. Because of the csc4Θ2\csc ^{4}\frac{\Theta }{2}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

                                P(Θ0)=1Q(Θ0)P(\Theta _{0})=1-Q(\Theta _{0})

                                where Q (θO) is the fraction of particles scattered through ΘQ(Θ0)\Theta\geq Q(\Theta _{0})

                                This fraction has been calculated before and is

                                Q(Θ0)=πn[Ze2(4πε0)T)2]cot2Θ02Q(\Theta _{0})=\pi n\left [ \frac{Ze^{2}}{(4\pi \varepsilon _{0})T})^{2}\right ]\cot^{2}\frac{\Theta _{0}}{2}

                                where n here is number of nuclei/cm2 .

                                Using the data we get Q = 0.4

                                P(Θo)=6P(\Theta _{o})=6


                              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°

                                Solution:

                                1. We know that

                                  The relevant fraction can be immediately written down

                                  ΔNN=[e2(4πε0)2T]2πcot2θo2.(n1.Z12n2.Z22)\frac{\Delta N}{N}=\left [ \frac{e^{2}}{(4\pi \varepsilon _{0})2T} \right ]^{2}\pi \cot^{2}\frac{\theta _{o}}{2}.(n_{1}.Z_{1}^{2}n_{2}.Z_{2}^{2})

                                  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

                                  n1=ρdNAM1=0.7n_{1}=\frac{\rho dN_{A}}{M_{1}}=0.7

                                  n2=ρdNAM2=0.3n_{2}=\frac{\rho dN_{A}}{M_{2}}=0.3

                                  Here M1 , M 2 are the mass numbers of Zn and Cu. Then, substituting the values Z1 = 30, Z2 = 29, M1 = 65.4, M2 = 63.5, we get

                                  ΔNN=1.43×103\frac{\Delta N}{N}=1.43\times10^{-3}


                                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°

                                  Solution:

                                  1. From Rutherford scattering formula

                                    IE IRODOV Chapter 6 Question 16

                                    For Uranium nucleus Z - 92 and on substituting the values we get

                                    Δσ=737b=0.737kb\Delta \sigma =737b=0.737kb


                                  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°.

                                    Solution:

                                    1. (a)From the previous formula, we know that

                                      IE IRODOV Chapter 6 Question 17

                                      Substituting the values with Z = 79

                                      we get (θO = 90°)

                                      T = 0.903 MeV

                                      (b) The differential scattering cross section is

                                      IE IRODOV Chapter 6 Question 17 Solution


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

                                    dEdT=2e23e2w2\frac{dE}{dT}=\frac{-2e^{2}}{-3e^{2}}w^{2}

                                    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.

                                      Solution:

                                      1. The formula in MKS units is

                                        dEdT=μ0e26πcw2\frac{dE}{dT}=\frac{\mu _{0}e^{2}}{6\pi c}\vec{w^{2}}

                                        For an electron performing (linear) harmonic vibrations w\vec{w} is in some definite directions with

                                        Assume wx=w2x

                                        Thus,

                                        dEdT=μ0e2ω46πcx2\frac{dE}{dT}=\frac{\mu _{0}e^{2}\omega ^{4}}{6\pi c}{x^{2}}

                                        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

                                        E=12mω2a2E=\frac{1}{2}m\omega ^{2}a^{2}

                                        x=acosωtx=a\cos\omega t

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

                                        IE IRODOV Chapter 6 Question 18


                                      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.

                                        Solution:

                                        1. 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

                                          IE IRODOV Chapter 6 Question 19

                                          On integrating the above equation we get

                                          IE IRODOV Chapter 6 Question 19 Solution


                                        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 → ∞ the frequency of the photon ω → ωn.

                                          Solution:

                                          1. In a circular orbit we have the following formula

                                            IE IRODOV Chapter 6 Question 20

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

                                            IE IRODOV Chapter 6 Question 20 Solutions

                                            The above equation can also be written as

                                            IE IRODOV Chapter 6.1 Solved Question 20