Q: 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.

tanθ/2=(eπε0q1q2)/2bT Thus b=(eπε0q1q2)2Tcosθ/2 For α particle q1=2e for gold q2=79e (In Gaussian units there is no factor (4πε01) )Substituting we get b = 0.731 pm .

Q: Q.7. A point source is suspended at a height h = 1.0 m over the centre of a round table of radius R = 1.0 m. The luminous intensity I of the source depends on direction so that illuminance at all points of the table is the same. Find the function I (θ), where θ is the angle between the radiation direction and the vertical, as well as the luminous flux reaching the table if I (0) = I0 = 100 cd.

The illuminance at P is Ep=(x2+h2)I(θ)cosθ=h2I(θ)cos3θ since this is constant at all x, we must haveI(θ)cos3θ= const=I0 or I(θ)=I0/cos3θ The luminous flux reaching the table isΦ=πR2×h2I0=314 lumen

Q: 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 hemisphereθ<2π

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 θ=π−2φ or φ=2π−2θAlsob=(R+r)sinφ=(R+r)cos2θ.(b) With b defined above, the fraction of particles scattered between θ and θ+dθ (or the probability of the same) isdP=π(R+r)2∣2πbdb∣=21sinθdθ(c) This isP=∫0π/221sin0dθ=21∫−10d(−cosθ)=21

Question 1

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.

Accepted Answer

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 nucleus is +e while the charge on the electron is -e. The electron by symmetry must be at the centre of the nuclear charge where the potential isφ0​=(4πε0​1​)2R3e​ where R is the radius of the nuclear charge distribution. The potential energy of the electron is −eφ0​  and since the electron is at rest, this is also the total energy. To ionize such an electron will require an energy of  E=eφ0​ From this we find  R=(4πε0​1​)2E3e2​ In Gaussian system the factor 4πε0​1​  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 energy δE of oscillation then since the potential at a distance r in the nucleus isφ(r)=(4πε0​1​)Re​(23​−2R2r2​) the total energy of the nucleus becomes21​mr2−(4πε0​1​)Re2​(23​−2R2r2​)=−eφ0​+δE or δE=21​mr˙2+(4πε0​1​)2R3e2​r2 This is the energy of a harmonic oscillator whose frequency is ω2=(4πε0​1​)mR3e2​ The vibrating electron emits radiation of frequency ω whose wavelength isλ=ω2πc​=e2πc​mR3​(4πε0​)1/2 In Gaussian units the factor  (4πε0​)1/2   is missing.Putting the values we get λ=0.237μm

Question 2

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.

Accepted Answer

tanθ/2=(eπε0​q1​q2​​)/2bT Thus b=(eπε0​q1​q2​​)2Tcosθ/2​ For α particle q1​=2e for gold q2​=79e (In Gaussian units there is no factor (4πε0​1​) )Substituting we get          b = 0.731 pm .

Question 3

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 Li^z nucleus?

Accepted Answer

Ans.  (a) In the Pb case we shall ignore the recoil of the nucleus both because Pb is quite heavy (APb​=208=52×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(4πε0​)rmin​2Ze2​=T (No (4πε0​)  in Gaussian units.). Thus putting the valuesrmin = 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, x2 = coordinate of the Li nucleus with respect to the same point Then we have the energy momentum equations21​m1​x˙12​+21​m2​x˙22​+(4πε0​)∣x1​−x2​∣2×3e2​=Tm1​x˙1​+m2​x˙2​=2m1​T​Herc m1​= mass of He++nucleus, m2​= mass of Li nucleus. Eliminating x˙2​T=21​m1​x˙12​+2m2​1​(2m1​T​−m1​x˙1​)2+(4πε0​)(x1​−x2​)6e2​We complete the square on the right hand side and rewrite the above equation asm1​+m2​m2​​T=2m2​1​[m1​(m1​+m2​)​x˙1​−m1​+m2​m1​​​2m1​T​]2+(4πε0​)∣x1​−x2​∣6e2​For the least distance of apporach, the second term on the right must be greatest which implies that the first term must vanish.Thus ∣x1​−x2​∣min​=(4πε0​)T6e2​(1+m2​m1​​)Using m2​m1​​=74​and other values we get∣x1​−x2​∣min​=.034pm. (In Gaussian units the factor 4πε0​ is absent).

Question 4

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.

Accepted Answer

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 v0 is the speed of the particle at A, the following equations holdΓ=21​mv02​+(4πε0​)rmin​z1​z2​e2​ (1)mv0​rmin​=2mT​bρmin​mv02​​=(4πε0​)rmin2​z1​z2​e2​(This is Newton's law. Here ρ=ρmin​ 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 formb=(4πε0​)2TZ1​Z2​e2​cot2θ​From (2) and (3) ρmin​2Tb2​=(4πε0​)Z1​Z2​e2​or ρmin​=(4πε0​)2Tz1​z2​e2​cot22θ​,z1​=2,z2​=80 we get with ρmin​=0⋅231pm.(b) From (2) and (4) we writermin​=(4πε0​)2mT​Z1​Z2​e2​v0​cotθ/2​,Substituting in (1)T=21​mv02​+2mT​v0​tanθ/2v0​=m2T​​(sec2θ​−tan2θ​)Solving for v0​ we getThen rmin​=(4πε0​)2Tz1​z2​e2​sec2θ​−tan2θ​cot2θ​​=(4πε0​)2Tz1​z2​e2​cot2θ​(sec2θ​+tan2θ​)=(4πε0​)2Tz1​z2​e2​(1+cosec2θ​)=0.557 pm

Question 5

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.

Accepted Answer

By momentum conservationThus the momentum transferred td the gold nucleus is clearlyΔP=Pf​−Pi​=p​−p​ 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. ThenΔP=2mT​(1−cosθ)i^2mT​sinθj^​Here i^is  the unit vector in the direction of the incident proton and ∣j^​ is normal to it on the side on which it is scattered. Thus ∣ΔP∣∼22mT​sin2θ​Or using tanθ/2=(4πε0​)2bTZe2​ for the proton we get∣ΔP∣∼22mT/(1+(Ze22bT(4πε0​)​)2)​

Question 6

Q.6. An illuminant shaped as a plane horizontal disc S = 100 cm^2 in area is suspended over the centre of a round table of radius R = 1.0 m. Its luminance does not depend on direction and is equal to L = 1.6.10^4  cd/m^2. At what height over the table should the illuminant be suspended to provide maximum illuminance at the circumference of the table? How great will that illuminance be? The illuminant is assumed to be a point source.

Accepted Answer

The equivalent luminous intensity in the direction OP  is LScosθ (R2+h2)LScosθ​cosθ=(R2+h2)2LSh2​ =(hR2​+h)2LS​=[(h​R​−h​)2+2R]2LS​ This is maximum when and the maximum illuminance is        R = h4R2LS​=41.6×102​=40 lux

Question 7

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.

Accepted Answer

The proton moving by the electron first accelerates and then decelerates 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 OAand its magnitude isPd​=∫F⊥​dtwhere F⊥​ is the component along OA of the force on electron. ThusPd​=∫−∞∞​4πε0​e2​b2+v2t2​b​⋅b2+v2t21​dt=4πε0​ve2b​⋅∫−∞∞​(b2+x2)3/2dx​Evaluate the integral by substitutingx−5tan8Then Pe​=(4πε0​)vb2e2​Te​=2me​Pe2​​=(4πε0​)2Tb2me​mp​e4​ThenIn Gaussian units there is no factor (4πe0​)2. Substituting the values we get Te​=3.82eV.

Question 8

Q.7. A point source is suspended at a height h = 1.0 m over the centre of a round table of radius R = 1.0 m. The luminous intensity I of the source depends on direction so that illuminance at all points of the table is the same. Find the function I (θ), where θ is the angle between the radiation direction and the vertical, as well as the luminous flux reaching the table if I (0) = I0 = 100 cd.

Accepted Answer

The illuminance at P is Ep​=(x2+h2)I(θ)​cosθ=h2I(θ)cos3θ​ since this is constant at all x, we must haveI(θ)cos3θ= const=I0​ or I(θ)=I0​/cos3θ  The luminous flux reaching the table isΦ=πR2×h2I0​​=314 lumen

Question 9

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.

Accepted Answer

See the diagram on the next page. 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​​ . . 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+U0​T​​sinα=nsinα​ where n=1+TU0​​​   We also haveθ=2(α−φ) Therefore sin2θ​=sin(α−φ)=sinαcosφ−cosαsinφ=sinα(cosφ−ncosα​)or sinαnsinθ/2​=n2−sin2α−cosα​or (sinαnsinθ/2​+cosα)2=n2−sin2αor n2sin22θ​cot2α+2nsin2θ​cotα+1=n2cos22θ​or cotα=nsin2θ​ncos2θ​−1​ Hence sinα=1+n2−2ncos2θ​​nsin2θ​​Finally, the impact parameter isb=Rsinα=1+n2−2ncos2θ​​nRsin2θ​​

Question 10

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 hemisphereθ<2π​

Accepted Answer

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  θ=π−2φ or φ=2π​−2θ​Alsob=(R+r)sinφ=(R+r)cos2θ​.(b) With b defined above, the fraction of particles scattered between θ and θ+dθ (or the probability of the same) isdP=π(R+r)2∣2πbdb∣​=21​sinθdθ(c) This isP=∫0π/2​21​sin0dθ=21​∫−10​d(−cosθ)=21​

Question 11

Q.8. A vertical shaft of light from a projector forms a light spot S = 100 cm^2  in area on the ceiling of a round room of radius R = = 2.0 m. The illuminance of the spot is equal to E = 1000 lx. The reflection coefficient of the ceiling is equal to p = 0.80. Find the maximum illuminance of the wall produced by the light reflected from the ceiling. The reflection is assumed to obey Lambert's law.

Accepted Answer

The illuminated area acts as a Lambert source of luminosity M = πL where  MS=ρES= total reflected lightThus, the luminanceL=πρE​ The equivalent luminous intensity in the direction making an angle θ from the vertical isLScosθ=πρES​cosθ and the illuminance at the point P isπρES​cosθsinθ/R2cosec2θ=πR2ρES​cosθsin3θ This is maximum whendθd​(cosθsin3θ)=−sin4θ+3sin2θcos2θ=0 or tan2θ=3⇒tanθ=3​ Then the maximum illuminance is16π33​​R2ρES​ This illuminance is obtained at a distance Rcotθ=R/3​ from the ceiling. Substitution gives the value0.21 lux

Question 12

Q.9. A luminous dome shaped as a hemisphere rests on a horizontal plane. Its luminosity is uniform. Determine the illuminance at the centre of that plane if its luminance equals L and is independent of direction.

Accepted Answer

From the definition of luminance, the energy emitted in the radial direction by an element dS of the surface of the dome is dΦ=LdSdΩ Here L= constant. The solid angle dΩ is given bydΩ=R2dAcosθ​ where dA is the area of an element on the plane illuminated by the radial light. ThendΦ=R2LdSdA​cosθ The illuminance at 0 is thenE=∫dAdΦ​=∫0π/2​R2L​2πR2sinθdθcosθ=2πL∫0t​xdx=πL

Question 13

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 cm^2  located at the distance 10 cm from the scattering section of the foil. What fraction of scattered alpha particles reaches the counter inlet?

Accepted Answer

NdN​=n((4πε0​)2TZe2​)2sin42θ​dΩ​ We have put q1​=2e,q2​=Ze  here. Also n = no. of Pt nuclei in the foil per unit areaUsing the values APt​=195,ρ=21.5×103 kg/m3 NA​=6.023×1026 /kilo molewe get  n=6.641×1022 per m2 Also dΩ=r2dSn​​=10−2 Sr Substituting we get  NdN​=3.36×10−5

Question 14

Q.10. A Lambert source has the form of an infinite plane. Its luminance is equal to L. Find the illuminance of an area element oriented parallel to the given source.

Accepted Answer

Consider an element of area dS at point P.It emits light of flux dΦ=LdSdΩcosθ =LdSh2sec2θdA​⋅cos2θ =h2LdSdA​cos4θ in the direction of the surface element dA at O.The total illuminance at O is thenE=∫h2LdS​cos4θ But dS=2πrdr=2πhtanθd(htanθ) =2πh2sec2θtanθdθ Substitution gives E=2πL∫0π/2​sinθcosθdθ=πL

Question 15

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.

Accepted Answer

From the formula (6.1 b) of the book, we finddNAg​dNPT​​=nAg​nPt​​⋅ZAg2​ZPT2​​=η But since the foils have the same mass thickness ( = pd ) , we havenAg​nPt​​=APt​AAg​​ see the problem (6.9). HenceZPt​=ZAg​⋅APt​ηAAg​​​ Substituting ZAg​=47,AAg​=108,APt​=195 and η=1.52  we getZPt​=77.86≈78

Question 16

Q.11. An illuminant shaped as a plane horizontal disc of radius R = 25 cm is suspended over a table at a height h = 75 cm. The illuminance of the table below the centre of the illuminant is equal to E0  = 70 lx. Assuming the source to obey Lambert's law, find its luminosity.

Accepted Answer

Consider an angular element of area 2πxdx=2πh2tanθsec2θdθ Light emitted from this ring isdΦ=LdΩ(2πh2tanθsec2θdθ)⋅cosθ Now,    dΩ=h2sec2θdAcosθ​ where dA = an element of area of the table just below the untre of the illuminant Then the illuminance at the element dA will beE0​=∫θ=0θ=α​2πLsinθcosθdθ where sinα=h2+R2​R​. Finally using luminosity M = πLE0​=Msin2α=Mh2+R2R2​ or M=E0​(1+R2h2​)=700 lm/m2 (1lx=1 m2 lm​ dimensionally)

Question 17

Q.12. A small lamp having the form of a uniformly luminous sphere of radius R = 6.0 cm is suspended at a height h = 3.0 m above the floor. The luminance of the lamp is equal to L = 2.0.10^4  cd/m^2 and is independent of direction. Find the illuminance of the floor directly below the lamp.

Accepted Answer

See the figure below. The light emitted by an element of the illuminant towards the point 0 under consideration is dΦ=LdSdΩcos(α+β) The element dS has the areadS=2πR2sinαdα cos(α+β)=OAhcosα−R​ The distanceOA=[h2+R2−2hRcosα]1/2 we also havesinαOA​=sin(α+β)h​=sinβR​ From the diagramcos(α+β)=OAhcosα−R​ cosβ=OAh−Rcosα​ If we imagine a small area dΣ at O thenOA2dΣcosβ​=dΩ Hence, the illuminance at O is∫dΣdΦ​=∫L2πR2sinαdα(OA)4(hcosα−R)(h−Rcosα)​ The limit of α is α=0 to that value for which α+β=90∘ , for then light is emitted tangentially. Thus,αmax​=cos−1hR​ Thus, E=∫0cos−1hR​​L⋅2πR2sinαdα(h2+R2−2hRcosα)2(h−Rcosα)(hcosα−R)​ we put, y=h2+R2−2hRcosα So, dy=2hRsinαdα E=∫(h−R)2h2−R2​L⋅R22hRdy​y2(h−2hh2+R2−y​)(2Rh2+R2−y​−R)​ =8h2R2L⋅2πR2​∫(h−R)2h2−R2​y2(h2−R2+y)(h2−R2−y)​dy =4h2πL​∫(h−R)2h2−R2​[y2(h2−R2)2​−1]dy=4h2πL​[−y(h2−R2)2​−y](h−R)2h2−R2​ =4h2πL​∫(h−R)2h2−R2​[y2(h2−R2)2​−1]dy=4h2πL​[−y(h2−R2)2​−y](h−R)2h2−R2​ =4h2πL​[(h+R)2−(h2−R2)−(h2−R2)+(h−R)2] =4h2πL​[2h2+2R2−2h2+2R2]=h2πLR2​ Substitution gives : E=25.1 lux

Question 18

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/cm^2. The beam intensity is I0  = 5.0.10^5  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°.

Accepted Answer

(a) From Eqn. (6.1 b) we get  dN=I0​τAAu​ρ⋅d⋅NA​​((4πε0​)2TZe2​)sin4θ/22πsinθdθ​(we have used dΩ=2πsinθdθ and N=I0​I )From the datadθ=2∘=57.32​radianAlso ZAu​=79,AAu​=197. Putting the values we getdN=1.63×10∧6(b) This number isN(θ0​)=I0​τ(AAu​ρdNA​​)((4πε0​)2TZe2​)24π∫θ0​π​sin32θ​cos2θ​dθ​The integral is2∫sin2θ0​​1​x2dx​=21​[2x2−1​]sin2θ0​​1​=cot22θ0​​ThusN(θ0​)=πnd((4πε0​)TZe2​)2I0​τcot22θ0​​where n is the concentration of nuclei in the foil. (n=ρN1​/AAAu​​)Substitution givesN(θ0​)=2.02×107

Question 19

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°).

Accepted Answer

The requisite probability can be written easily by analogy with (b) of the previous problem. It is P=I0​τN(π/2)​=nd((4πε0​)2mv2Ze2​)24π∫π/2π​sin32θ​cosθ/2dθ​The integral is unity. ThusP=πnd((4πε0​)mv2Ze2​)2Substitution gives usingn=AAg​ρAx​NA​​=10810.5×103×6.023×1026​,P=.006

Question 20

Q.13. Write the law of reflection of a light beam from a mirror in vector form, using the directing unit vectors e and e' of the inci- dent and reflected beams and the unit vector n of the outside normal to the mirror surface.

Accepted Answer

We see from the diagram that because of the law of reflection, the component of the incident unit  vector e along n changes sign on reflection while the component ∥ to the mirror remains unchanged.Writing e=e∥​​+e⊥​​ e∥​​=e−n(e⋅n) we see that the reflected unit vector ise=ei​​−e⊥​​=e−2n(e⋅n)

Chapter 6 : PART SIX. ATOMIC AND NUCLEAR PHYSICS