1

Physics Olympiad for P.G. students - 2019

Solutions to Multiple Choice Questions

PART A

1. Ans: (c)

∇ =∂

∂xi+

∂

∂yj +

∂

∂zk

∂

∂x(x2 + y2 + z2)3/2 =

3

2(x2 + y2 + z2)1/22x = 3rx

∇r3 = 3r(xi+ yj + zk) = 3rr.

2. Ans:(c)Since a, b and c are mutually orthogonal,

a · c = 0 =⇒ λ− 1 + 2µ = 0.

b · c = 0 =⇒ 2λ+ 4 + µ = 0.

Solving, we get λ = −3, µ = 2.

3. Ans: (c)This is a problem in combination. You do not worry about the order in whichyou choose the 11 players. The number of teams, you can form is 15C11.

15C11 =15!

11! 4!= 1365.

4. Ans: (c)

E2 = p2c2 +m20c

4 =⇒ 2EdE = 2pdp c2 =⇒ v =dE

dp=

p

Ec2

p2c2 = E2 −m20c

4 =⇒ pc = E

(1− m2

0c4

E2

)1/2

v =pc2

E= c

(1− m2

0c4

E2

)1/2

.

5. Ans: (d) [0 11 0

]ψ = λ ψ =⇒

[−λ 1

1 −λ

]ψ = 0.

2

For this to have a non-trivial solution,∣∣∣∣∣ −λ 11 −λ

∣∣∣∣∣ = 0 =⇒ λ2 = 1 or λ = ±1.

6. Ans: (b)r is a vector and σ is a pseudo-vector. Hence σ · r is a pseudo-scalar.

7. Ans: (b) ∮c

3z2 + z

z2 − 1dz =

∮c

3z2 + z

(z + 1)(z − 1)dz

The integral has simple poles at z = −1 and z = +1∮c

3z2 + z

z2 − 1dz = 2πiR1 +R2

R1 = Residue at z = −1

= limz→−1

(z + 1)(3z2 + z)

z2 − 1= lim

z→−1

3z2 + z

z − 1=

3− 1

−2= −1

R2 = Residue at z = +1

= limz→+1

(z − 1)(3z2 + z)

z2 − 1= lim

z→+1

3z2 + z

z + 1=

4

2= 2

Hence ∮c

3z2 + z

z2 − 1dz = 2πi−1 + 2 = 2πi.

8. Ans: (d)de Broglie wavelength λ = h

p.

Since the kinetic energy E = p2

2m, p = (2mE)1/2. Therefore λ1

λ2=(m2

m1

)1/2.

9. Ans: (b)Since L = r × p, Ly = −xpz + pxz.p2x will commute with everything except x. So, the commutator

[p2x, Ly]− = −[p2

x, x]−pz = 2ihpxpz,

since the commutator

[p2x, x]− = px[px, x]− + [px, x]−px = −2ihpx.

The correct answer is (b).

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10. Ans: (a)For any point outside the charged sphere, it will appear as if the entire chargeis concentrated at its centre. So, electrostatic force F between the two spheresis

F =1

4πε0

q1q2

r2,

where q1 = q2 = 6.5× 10−7 Coulomb;r = 0.5 metre;1/4πε0 = 9× 109 N m2 C−2.Substituting these values, we get F = 15.21× 10−3 N.

11. Ans: (d)If N is the number of molecules in volume V , then V = N/ρ and dV =−(N/ρ2)dρ.Work done in isothermal compression

dW = −∫PdV =

∫(aρ+ bρ2)(N/ρ2)dρ

= N∫ 2ρ0

ρ0

(a

ρ+ b

)dρ = N (a ln ρ+ bρ) |2ρ0ρ0

= N (a ln 2 + bρ0) = ρ0V0 (a ln 2 + bρ0)

12. Ans: (a)

x2 + 4y2 = 8 =⇒ 2xdx

dt+ 8y

dy

dt= 0.

At x = 2, y = 1, 4dxdt

= −8dydt

. Since it is given that dxdt

= 6, it follows thatdydt

= −3.

13. Ans: (a)

V =1

2k(t)x2 =⇒ 〈V 〉 =

1

2k(t)〈x2〉

d

dt〈V 〉 =

1

2

dk

dt〈x2〉+

k

2

⟨xdx

dt+dx

dtx

⟩

=1

2

dk

dt〈x2〉+

k

2m〈xp+ px〉

Alteranative method

Applying Ehrenfest theorem,

d

dt〈A〉 =

⟨∂A

∂t

⟩+

1

ih〈[A,H]〉 ,

4

we get

d

dt〈V 〉 =

⟨∂V

∂t

⟩+

1

ih〈[V,H]〉 =

1

2

dk

dt〈x2〉+

1

ih

⟨[V,

p2

2m+ V

]⟩,

Let us now evaluate the commutator

[V,H] =1

2kx2

(p2

2m+

1

2kx2

)−(p2

2m+

1

2kx2

)1

2kx2 =

1

2

k

2m[x2, p2]

=k

4m

(x[x, p2] + [x, p2]x

)=

k

4m2ih(xp+ px)

since [x, p2] = 2ihp. Hence

d

dt〈V 〉 =

⟨∂V

∂t

⟩+

1

ih〈[V,H]〉 =

1

2

dk

dt〈x2〉+

k

2m〈xp+ px〉.

14. Ans: (b)Due to parity conservation in strong interaction, the two neutrons in the finalstate should be in p state (l = 1). The total wave function of two neutronsshould be antisymmetric.

Total wave function (n,n) = (orbital) (spin) (isospin)

= (antisymmetric) (symmetric) (symmetric)

Hence spin of n-n: S = 1 (Symmetric)

15. Ans: (b)Radius R is proportional to A1/3.

RAl

RZn

=(27)1/3

(64)1/3=

3

4.

So, RAl = 34× 4.8× 10−13 = 3.6× 10−13 cm.

16. Ans: (c)Nucleon configuration of the ground state and the first excited state of 17O8:

Ground state:(1s1/2

)2p,2n (1p3/2

)4p,4n (1p1/2

)2p,2n (1d5/2

)1n

First excited state:(1s1/2

)2p,2n (1p3/2

)4p,4n (1p1/2

)2p,2n (2s1/2

)1n

Spin-parity of Ground state of 17O8 = 5/2+.Spin-parity of First excited state of 17O8 = 1/2+.

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17. Ans: (a)The rotational energy levels of the nucleus

Ej =h2

2Ij(j + 1),

where I denotes the moment of inertia and j can take values 0, 2, 4, · · ·. Sincej = 0 corresponds to the ground state, j = 2 corresponds to the first excitedstate and j = 4 corresponds to the second excited state.

E2 =6h2

2I= 90 KeV; E4 =

20h2

2I= 300 KeV

18. Ans: (c)

px =∂L

∂x= 2ax; py =

∂L

∂y= 2by

H =∑i

piqi − L = pxx+ pyy − ax2 − by2 + kxy

= 2ax2 + 2by2 − ax2 − by2 + kxy = ax2 + by2 + kxy

=p2x

4a+p2y

4b+ kxy

19. Ans: (b)

t′B − t′A = γ

(tB − tA)− v

c2(xB − xA)

.

Data given: tB − tA = 0; xB − xA = 10× 103 m; v/c = 0.95;c = 3× 108 m.

γ =1√

1− (v2/c2)= 3.2026.

Substituting these values, we get t′B − t′A = −1.0142× 10−4 sec.From the negative sign, we infer that the flash at B occurs first and the flashat A happens after a time interval 1.0142× 10−4 sec.

20. Ans: (d)NaCl belongs to f.c.c lattice and so there are n = 4 molecules per unit cell.

a =

(nM

Na ρ

)1/3

=(

4× 58.45

6× 1026 × 2170

)1/3

= 5.64× 10−10 m

21. Ans: (c)

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22. Ans: (a)Arsenic, Antimony and Phosphorous are pentavalent impurities and Aluminiumis a trivalent impurity. Hence the intrinsic semiconductor doped with Alu-minium is a p-type semiconductor.

23. Ans: (c)

24. Ans: (b)

B = CurlA = ∇×A =

i j k∂∂x

∂∂y

∂∂z

Ax Ay Az

B = B0k =

(∂Ay∂x− ∂Ax

∂y

)k

Let Ay = 0. Then B0dy = −∂Ax or B0y = −Ax.This leads to a possible choice of the vector potential A = −B0yi.

25. Ans: (c)Units of h = JS, c = m/S, E = J . Unit of cross section is m2.

26. Ans: (b)Total effective resistance in the circuit = 50+70 = 120 Ω.Current through 50 Ω resistance = 12

120= 0.1 A.

Voltage across the resistance 50 Ω = 50× 0.1 = 5 Volts.

27. Ans: (a)

(A+ B

)· C =

(A+ B

)+ C, using De Morgan’s theorem

= (A · ¯B) + C, using De Morgan’s theorem

= (A ·B) + C

28. Ans: (c)The total wave function of two protons should be antisymmetric. Since thespin part is symmetric, the orbital part should be antisymmetric.

29. Ans: (b)It is a decay by weak interaction, for which ∆S = 1, 0 are allowed values. In(b) ∆S = 2, since S = −3 for Ω− and S = −1 for Λ0.

30. Ans: (c)

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PART B

31. Ans: (d)

tanh(x) =sinh(x)

cosh(x)=ex − e−x

ex + e−x=e2x − 1

e2x + 1=

1

y(e2x − 1

)y = e2x + 1 =⇒ e2x(y − 1) = y + 1 =⇒ e2x =

y + 1

y − 1

ex =

√y + 1

y − 1=⇒ x = ln

√y + 1

y − 1.

32. Ans: (a)In the stationary lift

T = 2π

√l

g.

When the lift accelerates upwards with an acceleration g/4,the effective accel-eration on the pendulum is g′ = g + g/4 = 5g/4.

T ′ = 2π

√l

g′= 2π

√4

5

l

g=

2T√5

33. Ans: (a)

F (k) =1

(2π)3/2

∫ ∞−∞

e−(x2/a2)+ikxxdx

∫ ∞−∞

e−(y2/a2)+ikyydy

∫ ∞−∞

e−(z2/a2)+ikzzdz∫ ∞

−∞e−(x

2/a2)+ikxxdx =∫ ∞−∞

e−(xa+i kxa2 )

2

e−k2xa

2/4dx

= e−k2xa

2/4∫ ∞−∞

e−(xa+i kxa2 )

2

dx

Put y = xa

+ ikxa2

. Then dy = 1adx.

∫ ∞−∞

e−(xa+i kxa2 )

2

dx = a∫ ∞−∞

e−y2

dy = a√π

Similarly performing the other integrals and substituting these values, we get

F (k) =1

(2π)3/2a3(π)3/2e−k

2xa

2/4 =

(a√2

)3

e−k2a2/4.

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34. Ans: (b)

d2x

dt2− 3

dx

dt+ 2x = 0

(D2 − 3D + 2)x = 0, where D =d

dt.

If m are the roots, then

m2 − 3m+ 2 = 0 =⇒ (m− 1)(m− 2) = 0.

This yields the values m = 1 and m = 2 and the solution is

x(t) = Aet +Be2t,

where A and B are constants to be determined by the boundary conditionsx = 0 at t = 0 and x = 1 at t = 1.

A+B = 0; Ae+Be2 = 1.

This yields the values A = −B and A(e− e2) = 1 or A = −B = 1e−e2

At t = 2, x = Aet +Be2t =1

e− e2(e2 − e4) = e+ e2.

35. Ans: (d)Condition for the existence of a non-trivial solution:

det

1 1 11 2 32 b 2c

= 0 =⇒ (4c− 3b)− (2c− 6) + (b− 4) = 0.

This yields the condition b = c+ 1. Substituting this, we get 1 1 11 2 32 b 2b− 2

xyz

=

000

This yields the following set of equations:

x+ y + z = 0

x+ 2y + 3z = 0

2x+ by + (2b− 2)z = 0

9

This yields the values: x = z, y = −2x. Let x = 1, then y = −2, z = 1. Thisyields the un-normalized solution x

yz

=

1−2

1

.The normalized solution is: x

yz

=1√6

1−2

1

.36. Ans: (b)

For monoatomic gas, γ = 5/3 and for adiabatic process, TV γ−1 = constant.

T1(L1A)2/3 = T2(L2A)2/3 =⇒ T1

T2

=(L2

L1

)2/3

.

37. Ans: (c)The arrangement shown in figure is equivalent to two capacitors of capacitancesC1 and C2 connected in series.

C1 =K1ε0A

d/2; C2 =

K2ε0A

d/2.

If C is the net capacitance, then

1

C=

1

C1

+1

C2

=d

2K1ε0A+

d

2K2ε0A=

d

2ε0A

(K1 +K2

K1K2

).

C =2ε0A

d

(K1K2

K1 +K2

).

38. Ans: (c):

Kinetic energy of a particle: K = 12mv2. Its velocity is v =

√2Km.

If the particle carries a charge q and moves in a uniform magnetic field B,then it will describe a circle of radius R.

mv2

R= Bqv =⇒ mv

R= Bq.

This yields the relation for the velocities and kinetic energies of alpha particleand proton describing the circle of the same radius R, in the same magneticfield B.

mpvpR

= Bqp;mαvαR

= Bqα =⇒ vαvp

=qαmp

qpmα

=2

4=

1

2

10

Since qα = 2qp and mα = 4mp, we obtain vαvp

= 12. In terms of kinetic energies,

vαvp

=

(2Kα

mα

mp

2Kp

)1/2

=1

2

(Kα

Kp

)1/2

.

Since vαvp

= 12, it follows that Kα = Kp = 1 MeV.

39. Ans: (d)The speed of electromagnetic wave in a medium of relative permittivity εr andrelative permeability µr is given by

v =1

√εrε0µrµ0

=c

√εrµr

, since c =1

√ε0µ0

.

Substituting the values εr = 2.14, µr = 1.3, c = 3× 108, we get

v =3× 108

√2.14× 1.3

= 1.8× 108 m s−1.

40. Ans: (b)Fringe width β = λD

d. Substituting the values λ = 6000 × 10−10 m, D= 2 m,

d = 4× 10−3 m, we get β = 0.3 mm

41. Ans: (b)The decay constant (λ) is the reciprocal of mean life (τ).

λα =1

1620; λβ =

1

405; λ = λα + λβ.

λ is the total decay constant.

λ =1

1620+

1

405=

1

324per year.

Applying radioactive decay formula,

N = N0e−λt,

we can find the time taken for the three-fourth of the sample to decay orone-fourth of the sample remaining. N = N0/4.

N

N0

=1

4= e−λt =⇒ λt = ln 4 =⇒ t =

ln 4

λ=

2 ln 2

λ

Substituting the values, we get t = 2× 324× 0.693 = 449 years.

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42. Ans: (c)V (x) is minimum at the value of x at which dV

dx= 0.

V (x) = −1

2kx2 +

1

4λx4 =⇒ dV

dx= −kx+ λx3 = 0 =⇒ x2 =

k

λ.

V (x) is minimum at x = x0: x20 = k/λ and it corresponds to the equilibrium

point.Expanding V (x) around the point x = x0, we get

V (x) = V (x0) + (x− x0)dV

dx

∣∣∣∣∣x=x0

+(x− x0)2

2!

d2V

dx2

∣∣∣∣∣x=x0

+ · · ·

V (x)− V (x0) =(x− x0)2

2

d2V

dx2

∣∣∣∣∣x=x0

, sincedV

dx

∣∣∣∣∣x=x0

= 0.

d2V

dx2= −k + 3λx2;

d2V

dx2

∣∣∣∣∣x=x0

= −k + 3λk

λ= 2k.

This yields

V (x)− V (x0) =(x− x0)2

2(2k) =

1

2K(x− x0)2,

where K = 2k. This represents a simple harmonic motion about the point

x = x0 with angular frequency ω =√K/m =

√2km

since K = 2k. If ν is thefrequency of oscillations, then

ν =ω

2π=

1

2π

√2k

m

43. Ans: (a)

J ·A = JxAx + JyAy + JzAz

=1

2(J+ + J−)Ax +

1

2i(J+ − J−)Ay + JzAz

=1

2J+(Ax − iAy) +

1

2J−(Ax + iAy) + JzAz

Since the operator J+ alone will connect the initial state |j,m〉 to the finalstate |j,m+ 1〉 yielding the matrix element

〈j,m+ 1|J+|j,m〉 = (j −m)(j +m+ 1)1/2h,

it follows that

〈j,m+ 1|J ·A|j,m〉 = 12(Ax − iAy)(j −m)(j +m+ 1)1/2h.

So, the correct answer is: (a).

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44. Ans: (a)The particle is in a mixture of states with quantum numbers n = 2 and n = 4,since the normalized wave function and energy of a particle in a definite staten in an infinite potential well of width L is given by

ψn =

√2

Lsin

nπx

L. En =

n2π2h2

2mL2=

n2h2

8mL2( Note: h = h/(2π))

The given normalized wave function of the particle is

ψ(x) =

√2

L

3

5sin

2πx

L+

4

5sin

4πx

L

=

3

5ψ2 +

4

5ψ4.

The particle is in a mixture of states with n = 2 and n = 4 with probabilities(35

)2and

(45

)2.Thus the energy of the particle is

E =9

25E2 +

16

25E4 =

9

25

(h2

2mL2

)+

16

25

(2h2

mL2

)=

73

50

(h2

mL2

).

45. Ans: (a)The force between two long parallel wires, each of length `, carrying currentsI1 and I2 in the same direction but separated by a distance r is given by Eq.

F =µ0I1I2`

2πr.

This is a force of attraction.If three long parallel conductors A,B,Ccarrying current in the same direction asshown in figure, then the middle conduc-tor B experiences an attractive force dueto conductors A and C, placed on eitherside. If F1 and F2 represent respectivelythe two forces that the conductor C experi-ences due to conductors A and C on eitherside, the net force acting on C is

F = F2 − F1 =µ0IBIC`

2πrBC− µ0IaIB`

2πrAB

=µ0`

2π

IBICrBC

− IAIBrAB

.

F2F1

rBCrAB

IA IB IC

A B C

Fig.

13

Substituting the values, IA = 2 Amp, IB = 3 Amp, IC =4 Amp, ` = 10 metres,rAB = rBC = 10 cm =0.1 m and µ0 = 4π × 10−7 henry/metre, we get

F =2× 10−7 × 10

0.1(12− 6) = 1200× 10−7 = 1.2× 10−4N

46 . Ans: (b)According to Gauss’ law in electrostatics, the total normal electric flux φ over aclosed surface depends on the total charge q residing within the closed surface.

φ =q

ε0,

where ε0 is the permittivity of the medium. Let q be the charge residing withina sphere of radius R/2 and Q be the charge residing in the whole solid sphereof radius R.

q = 4π∫ R/2

0ρ(r)r2dr = 4πρ0

∫ R/2

0

(1− ar

R

)r2dr

= 4πρ0

∣∣∣∣∣r3

3− a

R

r4

4

∣∣∣∣∣R/2

0

= 4πρ0R3(

1

24− a

64

)

= 4πε0R2

4ER/2, by Gauss’ law

So

ER/2 =ρ0

ε04R

(1

24− a

64

)

Q = 4π∫ R

0ρ(r)r2dr = 4πρ0

∫ R

0

(1− ar

R

)r2dr

= 4πρ0

∣∣∣∣∣r3

3− a

R

r4

4

∣∣∣∣∣R

0

= 4πρ0R3(

1

3− a

4

)= 4πε0R

2ER, by Gauss’ law

This yields

ER =ρ0

ε0R(

1

3− a

4

)It is given that ER/2 = 5

4ER.. Hence

4(

1

24− a

64

)=

5

4

(1

3− a

4

)This yields the value a = 1.

14

47. Ans: (a):The magnetic induction along the axis of the solenoid is given by

B = µ0nI,

where n denotes the number of turns per unit length and I the current passingthrough the solenoid. Substituting the values n = 1000/2 = 500, I = 5 Ampsand µ0 = 4π × 10−7 henry/metre, we get

B = 3.14× 10−3 Tesla.

48. Ans: (c):Let us consider the collision of two particles a and b resulting in a final statec of an aggregate of particles. a+ b→ c.In the present case, the four-momentum of the final state Pc in c.m. systemfor the threshold production of ΛΛ is

Pc = (2MΛ, 0).

The conservation of four-momentum leads to the relation

(Pa + Pb)2 = P2

c

P2a + P2

b + 2Pa ·Pb = P2c

M2p +M2

p + 2Pa ·Pb = (2MΛ)2

2Pa ·Pb = 4M2Λ − 2M2

p .

In the laboratory system, the target is at rest. So,

Pa = (Ea, pa); Pb = (Mp, 0).

Substituting these values of four-momenta, we get

2Pa ·Pb = 2EaMp = 4M2Λ − 2M2

p .

Thus, we find the total energy of the incident proton required for thresholdproduction of ΛΛ pair production to be

Ea =1

Mp

(2M2Λ −M2

p ).

Substituting the values Mp = 938 MeV and MΛ = 1115 MeV, we get Ea=1712.8 MeV. This is the total energy including the rest mass energy of theincident proton. So the required kinetic energy of the incident proton in lab-oratory frame for ΛΛ pair production is

Tp = 1712.8− 938 = 774.8 MeV.

15

49. Ans: (a):Since the electrons in the betatron are accelerated to very high velocities, itis reasonable to assume v ≈ c. At such high velocities, it is justifiable toneglect the rest mass energy of the electron and assume that the momentumof electron p = mv ≈ E/c. Under these approximations

mv2

R= Bev; mv = RBe;

E

c= mv = RBe.

Thus, we obtain the energy of the emerging electron to be

E = RBec,

50. Ans: (b)First, let us derive an expression for the Fermi energy.Number of quantum states available for electrons in momentum range p andp + dp in volume V is V 4πp2dp/h3. Since 2 electrons, one with spin up andanother with spin down, can occupy a quantum state, the number of electronsin volume V in momentum range 0 to pf is

N = 2∫ pF

0

V 4πp2dp

h3=

8πV p3F

3h3.

If Ef is the Fermi energy, then pf = (2mEF )1/2.

N =8πV (2mEF )3/2

3h3=⇒ EF =

h2

2m

(3N

8πV

)2/3

.

Volume of unit cell = a3 = (5.34× 10−10)3

m3. For bcc solid, there are 2 atomsper unit cell. So, the number of atoms per unit volume (m3) is 2/a3. Since itis given that it is a monovalent solid, the number of valence electrons n is thesame as the number of atoms. So, the number of valence electrons per unitvolume is

N

V= n =

2

a3.

EF =h2

2m

(3n

8π

)2/3

=h2

2m

(3

4πa3

)2/3

=h2

2ma2

(3

4π

)2/3

Substituting the values h = 6.626 × 10−34 Js, m = 9.109 × 10−31 kg anda = 5.34× 10−10 m, we get EF in Joules. To get EF in eV, we need to divideit by 1.602× 10−19 since 1 eV = 1.602× 10−19 J.

EF =(

1

1.602× 10−19

)6.6262 × 10−68

2× 9.11× 10−31 × 5.342 × 10−20

(3

4π

)2/3

= 2.03 eV.

16

51. Ans: (d)Let us use the concept of phase-space. The number of quantum states availablefor particles with momentum lying between p and p+ dp and confined withina volume V is:

dN =V 4πp2dp

h3,

where h3 is the volume occupied by one quantum state in phase-space due touncertainty principle, h denoting the Planck’s constant. If m is the mass ofthe particle, then its kinetic energy ε = p2/(2m)

p2 = 2mε =⇒ pdp = mdε; p2dp = (2mε)1/2mdε =√

2m3/2ε1/2dε.

Since 2 electrons, one with spin up and another with spin down, can occupyeach quantum state, the total number of electrons in volume V is

N =8πV

h3

√2m3/2

∫ εF

0ε1/2dε =

8πV

3h3(2mεF )3/2.

The total energy of electrons in volume V is

E =8πV

h3

√2m3/2

∫ εF

0ε3/2dε =

8πV

5h3(2m)3/2(εF )5/2.

The average energy of an electron is

ε =E

N=

3

5εF .

52. Ans: (d)The radial function is normalized if the probability of finding the electron is 1in the entire range of radial distance lying between r = 0 and r =∞.∫ ∞

0|R21(r)|2r2dr =

1

24a5

∫ ∞0

r4e−r/adr =1

24a5

4!

(1/a)5= 1.

Probability of finding the electron at the radial distance r is

Pr =1

24a5r4e−r/a

At maximum probability, dprdr

= 0.

dPrdr

=1

24a5

d

dr

(r4e−r/a

)=

1

24a5

(4r3e−r/aa− 1

ar4e−r/a

)= 0

This yields the value r = 4a for maximum probability. So, r0 = 4a.

17

53. Ans: (a)Let us use the following definite integrals:∫ ∞

−∞e−ax

2

dx =

√π

a;

∫ ∞−∞

x2e−ax2

dx =1

2a

√π

a.

The Hamiltonian, trial wave function and the energy are as follows:

H = − h2

2m

d2

dx2− αδ(x); ψ(x) = Ae−bx

2

; E =〈ψ|H|ψ〉〈ψ|ψ〉

.

We have

〈ψ|ψ〉 = A2∫ ∞−∞

e−2bx2dx = A2

√π

2b

〈ψ|H|ψ〉 =

⟨ψ

∣∣∣∣∣− h2

2m

d2

dx2− αδ(x)

∣∣∣∣∣ψ⟩

= I1 + I2

dψ

dx= A

d

dxe−bx

2

= Ae−bx2

(−2bx)

d2ψ

dx2= −2Ab

d

dx

(xe−bx

2)

= −2Abe−bx

2

+ xe−bx2

(−2bx)

= −2Abe−bx2

(1− 2bx2)

I1 =

⟨ψ

∣∣∣∣∣− h2

2m

d2

dx2

∣∣∣∣∣ψ⟩

= A2 h2

2m

∫ ∞−∞

2be−2bx2(1− 2bx2)

dx

=A2h2

2m2b∫ ∞−∞

e−2bx2dx− 2b∫ ∞−∞

x2e−2bx2dx

=A2h2b

m

√π

2b− 2b

1

4b

√π

2b

=A2h2

2m

√πb

2=A2h2

4m

√2πb.

I2 = −α〈ψ|δ(x)|ψ〉 = −A2α∫ ∞−∞

e−2bx2δ(x)dx

= −A2α

E =〈ψ|H|ψ〉〈ψ|ψ〉

=A2(h2

4m

√2πb− α

)A2√

π2b

=

√2b

π

(h2

4m

√2πb− α

)

=h2

4m2b−

√2b

πα.

To find the value of the parameter b for which the energy is minimum, put

18

dEdb

= 0. The resulting value of b is obtained as

√2πb =

2mα

h2 .

Substituting this value of b in the expression for E, we get

E = −mα2

πh2 .

54. Ans: (c)

Note that∫ ∞−∞

e−α2x2x dx =

∫ ∞−∞

e−α2x2x3 dx = 0;

∫ ∞−∞

e−α2x2x2dx =

√π

2α3

〈1|x|0〉 = 〈0|x|1〉 =(

2

π

)1/2

α2∫ ∞−∞

e−α2x2x2dx =

(2

π

)1/2

α2

(√π

2α3

)

=1√2α

=

(h

2mω

)1/2

.

〈ψ|x|ψ〉 = c20〈0|x|0〉+ c2

1〈1|x|1〉+ 2c0c1〈0|x|1〉 = 0 + 0 + 2c0c1〈0|x|1〉 = 〈0|x|1〉,if c0 = c1 = 1√

2such that c2

0 + c21 = 1.

55. Ans: (b)Let us write down the four-momentum equation for the K−-decayK− → µ− + νµ.

PK = Pµ + Pν or PK −Pν = Pµ.

Squaring the latter form of the four-momentum equation, we get, in the restframe of K−,

P2K + P2

ν − 2PK ·Pν = P2µ

m2K + 0− 2mKEν = m2

µ

Eν =m2K −m2

µ

2mK

.

In deducing the above equation, we have used the relation that the four-momentum square is just equal to the rest mass square of the particle. Also,the scalar product of two four-momenta P1 · P2 = E1E2 − p1 · p2.. If themomentum of one of the particles is zero, the scalar product of three-momentaof the two particles becomes zero. Substituting the masses n energy unitsmK = 494 MeV, mµ = 106 MeV, we get Eν = 235.63 MeV. So, the correctanswer is (b).

19

56. Ans: (c)The total spin S = Se + Sp of electron - proton system can be either 1 0r 0.

S2 = (Se + Sp)2 = S2

e + S2p + 2Se · Sp

〈Se · Sp〉 =1

2

〈S2〉 − 〈S2

e〉 − 〈S2p〉

=1

2

S(S + 1)− 3

2− 3

2

h2

For S=1, 〈Se · Sp〉 = 12

(2− 3

2

)h2 = 1

4h2

For S=0, 〈Se · Sp〉 = 12

(−3

2

)h2 = −3

4h2

So, the energy splitting = ∆E = a

14h2 −

(−3

4h2)

= ah2.

57. Ans: (b)The first order correction to the energy is

∆E =〈ψ0|H ′|ψ0〉〈ψ0|ψ0〉

; where H ′ = bx2 and ψ0 =1√πa3

0

e−r/a0 .

It can be easily checked that ψ0 is a normalized wave function.

〈ψ0|ψ0〉 =1

πa30

∫ ∞0

e−2r/a0r2dr∫dΩ = 1;

∫dΩ =

∫ π

0sin θdθ

∫ 2π

0dφ

=4π

πa30

∫ ∞0

e−2r/a0r2dr =4

a30

(a0

2

)3 ∫ ∞0

e−xx2dx = 1

Above, we have put x = 2r/a0 and the integral∫∞0 e−xx2dx = 2!.

PerturbationH ′ = bx2 = b(r sin θ cosφ)2,

since x = r sin θ cosφ in spherical polar coordinates.We have∫ 2π

0cos2 φdφ =

1

2

∫ 2π

0(1− cos 2φ)dφ = π∫ π

0sin2 θ sin θdθ = −

∫(1− cos2 θ)d(cosθ) = −

∫ −1

1(1− x2)dx =

4

3

Therefore, perturbation energy is

∆E = 〈ψ0|bx2|ψ0〉 =b

πa30

∫e−2r/a0(r2 sin2 θ cos2 φ)r2drdΩ∫

sin2 θ cos2 φdΩ =4

3π.

20

∆E =b

πa30

4π

3

∫ ∞0

e−2r/a0r4dr

=4b

3a30

(a0

2

)5 ∫ ∞0

e−xx4dx

=4b

3a30

(a0

2

)5

(4!) = ba20

58. Ans: (d)The following data are required:Bohr magneton µB = 9.274× 10−24 JT−1

Planck’s constant h = 6.626× 10−34 JsVelocity of light c = 3× 108 m s−1

The splitting of spectral lines in normal Zeeman effect is given by

∆E = µBH =⇒ h∆ν = µBH

ν =c

λ=⇒ ∆ν = − c

λ2∆λ =⇒ ∆λ = −λ

2

c

µBHh

Substituting the values λ = 660 nm = 660× 10−9m and H = 0.3 T, we get

∆λ =(660)2 × 9.274× 0.3

3× 6.626× 10−16 m = 6× 10−12 m = 6 pm

59. Ans: (c)The normalized radial functions for hydrogen-like atom in the ground state(1s state) is given by

R1s(r) = 2(z

a0

)3/2

e−zr/a0 ,

from which one can write down the initial radial wave function ui of electronin 1s state of 3H atom and the final radial wave function vf of electron in 1sstate of 3He atom.

ui = 2(

1

a0

)3/2

e−r/a0 ;

vf = 2(

2

a0

)3/2

e−2r/a0 ;

Applying the sudden approximation to the sudden change in the atomic stateof the electron arising from beta decay, we find the transition probability.

〈vf |ui〉 =8√

2

a30

∫ ∞0

e−3r/a0r2dr.

21

The integral can be carried out easily.∫ ∞0

e−αrr2dr =2

α3.

The result is

〈vf |ui〉 =8√

2

a30

2

α3, with α =

3

a0

=16√

2

27= 0.838.

The probability of finding the final state in the ground state of 3He atom is

P = |〈vf |ui〉|2 = 0.8382 = 0.702.

60. Ans: (a)The commutator [σ · p, H]− can be evaluated as shown below:.

[σ · p, H]− = [σ · p, c(α · p+ βµ)]−= c[σ · p,α · p]−

= c

[σ · p 0

0 σ · p

] [0 σ · p

σ · p 0

]

−[

0 σ · pσ · p 0

] [σ · p 0

0 σ · p

]

= c

[0 (σ · p)2

(σ · p)2 0

]−[

0 (σ · p)2

(σ · p)2 0

]= 0.

61. Ans: (c)Let ψ1 and ψ2 denote positive energy states and ψ3 and ψ4 denote negativeenergy states. Then

(cα · p+ βmc2)ψ1,2 = |E|ψ1,2; (cα · p+ βmc2)ψ3,4 = −|E|ψ3,4

Λ+ψ1,2 = 1; Λ+ψ3,4 = 0; Λ−ψ1,2 = 0; Λ−ψ3,4 = 1.

Please also refer to Solved Problem (11.2) & (11.3); Pages 311 & 312 of Q.M.book by V. Devanathan .The correct answer is: (c).

62. Ans: (a)Conductivity σ = e(neµe + nhµh)In an intrinsic semiconductor, ne = nh = ni.

22

So, σ = nie(µe + µh).Substituting the values, ni = 3.6 × 1019, µe = 0.54. µh = 0.18 and e =1.6× 10−19, we get

σ = 3.6× 1019 × 1.6× 10−19(0.54 + 0.18) = 4.1472 Ω−1m−1.

63. Ans: (c)According to the Nuclear Shell Model, the nucleon configuration is:

27Al13 :(1s1/2

)2p,2n (1p3/2

)4p,4n (1p1/2

)2p,2n (1d5/2

)5p,6n

According to the Schmidt model, only the odd nucleon contributes to themagnetic moment.

Magnetic moment = jgl ± (gs − gl)

1

2l + 1

, j = l ± 1

2.

For the odd proton in the state 1d5/2, j = 5/2, l = 2.

Magnetic Moment =5

2

1 + (5.586− 1)

1

5

= 4.793µn.

64. Ans:(b)At equilibrium,Number of transitions from |0〉 to |1〉 = Number of transitions from 1〉 to |0〉.P0nW0→1 = P1W1→0 + P1nW1→0.Induced absorption= spontaneous emission + Induced emission, since therecannot be any spontaneous absorption.

65. Ans: (d)For (A,B) = (1,0), Y ′ = 1, Y ′′ = 0, Y ′′′ = 1, Y = 1.For (A,B) = (1,1), Y ′ = 0, Y ′′ = 1, Y ′′′ = 1, Y = 0.For (A,B) = (0,0), Y ′ = 1, Y ′′ = 1, Y ′′′ = 1, Y = 0.