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Q T-1

Particles from the Sun1

Page 1 of 2

(Total Marks: 10)

Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperature

s and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be , its luminosity (radiation energy emitted per unit time), distance, . Note: ∫ ∫ ∫ ( ( ( ) ) ) , its radius, , and the Earth-Sun

A Radiation from the sun : A1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, solar surface. , of the 0.3

The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, , is given by

where

is the frequency and

is the area of the surface normal to the direction of the incident radiation. , placed 0.3 0.2

Now, consider a solar cell which consists of a thin disc of semiconducting material of area, perpendicular to the direction of the Sun’ rays. A2 A3

Using the Wien approximation, express the total radiated solar power, , incident on the surface of the solar cell, in terms of , , , and the fundamental constants , , . Express the number of photons, the solar cell in terms of , , , , per unit time per unit frequency interval incident on the surface of , and the fundamental constants , , .

Th du r l f h l r ll h “b d ” f r y, . We assume the following model. Every photon of energy excites an electron across the band gap. This electron contributes an energy, , as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). A4 A5 A6 A7 A8 Define , , where . Express the useful output power of the cell, and the fundamental constants , , . . and should be clearly shown. What . Estimate the versus and . The values at ?

u

, in terms of

, ,

1.0 0.2 1.0 1.0 0.2

Express the efficiency, , of this solar cell in terms of Make a qualitative sketch of is the slope of at

Let be the value of for which value of within an accuracy of The band gap of pure silicon is value.

is maximum. Obtain the cubic equation that gives . Hence calculate . . Calculate the efficiency,

, of a silicon solar cell using this

1

Amol Dighe (TIFR), Anwesh Mazumdar (HBCSE-TIFR) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

Q T-1

Page 2 of 2

In the late nineteenth century, Kelvin and Helmholtz (KH) proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, , and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. A9 A10 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, , of the Sun at present, in terms of G, and . Estimate the maximum possible time, (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. The calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: Th “ l r ur ”, , produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. B1 Calculate the flux density, , of the number of neutrinos arriving at the Earth, in units of The energy released in the above reaction is . Assume that the energy radiated by the Sun is entirely due to this reaction. Travelling from the core of the Sun to the Earth, some of the electron neutrinos, , are converted to other types of neutrinos, . The efficiency of the detector for detecting is 1/6 of its efficiency for detecting . If there is no neutrino conversion, we expect to detect an average of neutrinos in a year. However, due to the conversion, an average of neutrinos ( and combined) are actually detected per year. B2 In terms of and , calculate what fraction, , of is converted to . 0.4 In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone. Assume that an electron knocked out by a neutrino loses energy at a constant rate of per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, , determine the energy imparted to this electron ( , n, and . (Assume the electron to r d by the neutrino, in terms of , be at rest before its interaction with the neutrino.) The fusion of H into He inside the Sun takes place in several steps. Nucleus of (rest mass, ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a nucleus (rest mass, < ) and emitting a . The corresponding nuclear reaction is: 0.6 0.3 0.5

B3

2.0

When a Be nucleus is at rest and absorbs an electron also at rest, the emitted neutrino has energy . However, the nuclei are in random thermal motion due to the temperature at the core of the Sun, and act as moving neutrino sources. As a result, the energy of emitted neutrinos fluctuates with a root mean square (rms) value . B4 If = , calculate the rms speed of the Be nuclei, , and hence estimate depends on the rms value of the component of velocity along the line of sight). . (Hint: 2.0

Theoretical Task 1 (T-1) : Solutions

Particles from the Sun1

1 of 7

Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also con?rm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be M = 2.00 × 1030 kg, its radius, R = 7.00 × 108 m, its luminosity (radiation energy emitted per unit time), L = 3.85 × 1026 W, and the Earth-Sun distance, d = 1.50 × 1011 m. Note: (i) (ii) (iii) xeax dx = x2 eax dx = x3 eax dx = 1 x ? 2 a a eax + constant eax + constant eax + constant

x2 2x 2 ? 2 + 3 a a a

6 x3 3x2 6x ? 2 + 3 ? 4 a a a a

A. Radiation from the Sun : (A1) Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, Ts , of the solar surface. Solution: Stefan’s law: L = (4πR2 )(σTs4 ) Ts = L 4πR2 σ

1/4

[0.3]

= 5.76 × 103 K

The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, u(ν ), is given by R2 2πh 3 ν exp(?hν/kB Ts ), u(ν ) = A 2 d c2 where A is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, A, placed perpendicular to the direction of the Sun’s rays. (A2) Using the Wien approximation, express the total power, Pin , incident on the surface of the solar cell, in terms of A, R , d , Ts and the fundamental constants c, h, kB .

Amol Dighe (TIFR), Anwesh Mazumdar (HBCSE-TIFR) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

1

[0.3]

Theoretical Task 1 (T-1) : Solutions

Solution: Pin =

0 ∞ ∞

2 of 7

u(ν )dν =

0

A dν =

R2 2πh 3 ν exp(?hν/kB Ts )dν d 2 c2 kB Ts dx. h

4 4 R2 R2 12πkB 2πkB 4 4 · 6 = T A T A s s c2 h3 d2 c2 h3 d2

Let x =

kB Ts hν . Then, ν = x kB Ts h 2πhAR2 (kB Ts )4 c2 d 2 h4

∞

Pin =

x3 e?x dx =

0

(A3) Express the number of photons, nγ (ν ), per unit time per unit frequency interval incident on the surface of the solar cell in terms of A, R , d , Ts ν and the fundamental constants c, h, kB . Solution: nγ (ν ) = u(ν ) hν R2 2π = A 2 2 ν 2 exp(?hν/kB Ts ) d c

[0.2]

The semiconducting material of the solar cell has a “band gap” of energy, Eg . We assume the following model. Every photon of energy E ≥ Eg excites an electron across the band gap. This electron contributes an energy, Eg , as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). (A4) De?ne xg = hνg /kB Ts where Eg = hνg . Express the useful output power of the cell, Pout , in terms of xg , A, R , d , Ts and the fundamental constants c, h, kB . Solution: The useful power output is the useful energy quantum per photon, Eg ≡ hνg , multiplied by the number of photons with energy, E ≥ Eg .

∞

[1.0]

Pout = hνg

νg

nγ (ν )dν R2 2π d 2 c2

2 ∞

= hνg A

ν 2 exp(?hν/kB Ts )dν

νg

R 2π = kB Ts xg A 2 2 d c =

4 2πkB c2 h3

kB Ts h

3

∞

x2 e?x dx

xg

R2 4 ? xg Ts A 2 xg (x2 g + 2xg + 2)e d

(A5) Express the ef?ciency, η , of this solar cell in terms of xg .

[0.2]

Theoretical Task 1 (T-1) : Solutions

Solution: Ef?ciency η = Pout xg 2 = (x + 2xg + 2)e?xg Pin 6 g

3 of 7

(A6) Make a qualitative sketch of η versus xg . The values at xg = 0 and xg → ∞ should be clearly shown. What is the slope of η (xg ) at xg = 0 and xg → ∞? Solution: 1 3 ?xg (x + 2x2 g + 2xg )e 6 g Put limiting values, η (0) = 0 η (∞) = 0. Since the polynomial has all positive coef?cients, it increases monotonically; the exponential function decreases monotonically. Therefore, η has only one maximum. η= dη 1 2 ?xg = (?x3 g + xg + 2xg + 2)e dxg 6 dη dxg η =

xg =0

[1.0]

1 3

dη dxg

=0

xg →∞

xg (A7) Let x0 be the value of xg for which η is maximum. Obtain the cubic equation that gives x0 . Estimate the value of x0 within an accuracy of ±0.25. Hence calculate η (x0 ). Solution: The maximum will be for dη 1 2 ? xg = (?x3 =0 g + xg + 2xg + 2)e d xg 6

2 ? p(xg ) ≡ x3 g ? xg ? 2xg ? 2 = 0

[1.0]

A Numerical Solution by the Bisection Method: Now, p(0) = ?2 p(1) = ?4 p(2) = ?2 p(3) = 10 ? 2 < x0 < 3 p(2.5) = 2.375 ? 2 < x0 < 2.5 p(2.25) = ?0.171 ? 2.25 < x0 < 2.5 The approximate value of xg where η is maximum is x0 = 2.27.

Theoretical Task 1 (T-1) : Solutions

Alternative methods leading to the same result are acceptable. η (2.27) = 0.457

4 of 7

(A8) The band gap of pure silicon is Eg = 1.11 eV. Calculate the ef?ciency, ηSi , of a silicon solar cell using this value. Solution: 1.11 × 1.60 × 10?19 = 2.23 1.38 × 10?23 × 5763 xg ηSi = (x2 + 2xg + 2)e?xg = 0.457 6 g xg =

[0.2]

In the late nineteenth century, Kelvin and Helmholtz (KH) proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, M , and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational energy through this slow contraction. (A9) Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, ?, of the Sun at present, in terms of G, M and R . Solution:

M

[0.3]

The total gravitational potential energy of the Sun: ? = ?

0

Gm dm r

3M For constant density, ρ = 4πR3

R

4 m = πr3 ρ 3 4πr2 ρ

dm = 4πr2 ρdr

?=?

0

G

4 3 πr ρ 3

dr 16π 2 Gρ2 R5 3 GM 2 =? =? r 3 5 5 R

(A10) Estimate the maximum possible time τKH (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Solution: τKH = τKH ?? L

[0.5]

3GM 2 = = 1.88 × 107 years 5R L

The τKH calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.

Theoretical Task 1 (T-1) : Solutions

B. Neutrinos from the Sun:

5 of 7

In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: 41 H ?→ 4 He + 2e+ + 2νe The “electron neutrinos”, νe , produced in this reaction may be taken to be massless. They escape the Sun and their detection on Earth con?rms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. (B1) Calculate the ?ux density, Φν , of the number of neutrinos arriving at the Earth, in units of m?2 s?1 . The energy released in the above reaction is ?E = 4.0 × 10?12 J. Assume that the energy radiated by the Sun is almost entirely due to this reaction. Solution: 4.0 × 10?12 J ? 2ν 3.85 × 1026 L × 2 = × 2 = 6.8 × 1014 m?2 s?1 . 2 11 2 ? 12 4πd δE 4π × (1.50 × 10 ) × 4.0 × 10

[0.6]

? Φν =

Travelling from the core of the Sun to the Earth, some of the electron neutrinos, νe , are converted to other types of neutrinos, νx . The ef?ciency of the detector for detecting νx is 1/6th of its ef?ciency for detecting νe . If there is no neutrino conversion, we expect to detect an average of N1 neutrinos in a year. However, due to the conversion, an average of N2 neutrinos (νe and νx combined) are actually detected per year. (B2) In terms of N1 and N2 , calculate what fraction, f , of νe is converted to νx . Solution: [0.4]

N1 Ne Nx N2

= N0 = N0 (1 ? f ) = N0 f /6 = Ne + Nx

OR (1 ? f )N1 + ?f = 6 5 f N1 = N2 6 N2 1? N1

Theoretical Task 1 (T-1) : Solutions

6 of 7

In order to detect neutrinos, large detectors ?lled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, n), this radiation, called Cherenkov radiation, is emitted in the shape of a cone. (B3) Assume that an electron knocked out by a neutrino loses energy at a constant rate of α per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time ?t, determine the energy imparted to this electron (Eimparted ) by the neutrino, in terms of α, ?t, n, me , c. (Assume the electron to be at rest before its interaction with the neutrino.) Solution: When the electron stops emitting Cherenkov radiation, its speed has reduced to vstop = c/n. Its total energy at this time is Estop = nme c2 =√ 2 n2 ? 1 1 ? vstop /c2 m e c2

[2.0]

The energy of the electron when it was knocked out is nme c2 Estart = α?t + √ n2 ? 1 Before interacting, the energy of the electron was equal to me c2 . Thus, the energy imparted by the neutrino is Eimparted = Estart ? me c2 = α?t + n √ ? 1 me c2 2 n ?1

The fusion of H into He inside the Sun takes place in several steps. Nucleus of 7 Be (rest mass, mBe ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a 7 Li nucleus (rest mass mLi < mBe ) and emitting a νe . The corresponding nuclear reaction is:

7

Be + e? ?→ 7 Li + νe .

When a Be nucleus (mBe = 11.65 × 10?27 kg) is at rest and absorbs an electron also at rest, the emitted neutrino has energy Eν = 1.44 × 10?13 J. However, the Be nuclei are in random thermal motion due to the temperature Tc at the core of the Sun, and act as moving neutrino sources. As a result, the energy of emitted neutrinos ?uctuates with a root mean square value ?Erms . (B4) If ?Erms = 5.54 × 10?17 J, calculate the rms speed of the Be nuclei, VBe and hence estimate Tc . (Hint: ?Erms depends on the rms value of the component of velocity along the line of sight.) [2.0]

Theoretical Task 1 (T-1) : Solutions

Solution:

7 of 7

Moving 7 Be nuclei give rise to Doppler effect for neutrinos. Since the fractional change in energy (?Erms /Eν ? 10?4 ) is small, the Doppler shift may be considered in the nonrelativistic limit (a relativistic treatment gives almost same answer). Taking the line of sight along the z -direction, ?Erms vz,rms = Eν c = 3.85 × 10?4 1 VBe =√ 3 c ? VBe = √ 3 × 3.85 × 10?4 × 3.00 × 108 m s?1 = 2.01 × 105 m s?1 .

The average temperature is obtained by equating the average kinetic energy to the thermal energy. 3 1 2 mBe VBe = kB Tc 2 2 ? Tc = 1.13 × 107 K

Q T-2

The Extremum Principle1

A The Extremum Principle in Mechanics Consider a horizontal frictionless plane shown in Fig. 1. It is divided into two regions, I and II, by a line AB satisfying the equation . The potential energy of a point particle of mass in region I is while it is in region II. The particle is sent from the origin O with speed along a line making an angle with the x-axis. It reaches point P in region II traveling with speed along a line that makes an angle with the x-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). A1 A2 Obtain an expression for in terms of Express in terms of , and . , and . I A II

Page 1 of 2

(Total Marks: 10)

×

B

Figure 1

0.2 0.3

We define a quantity called action , where is the infinitesimal length along the trajectory ∫ of a particle of mass moving with speed . The integral is taken over the path. As an example, for a particle moving with constant speed on a circular path of radius , the action for one revolution will be . For a particle with constant energy , it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). PLA implies that the trajectory of a particle moving between two fixed points in a region of constant potential will be a straight line. Let the two fixed points and in Fig. 1 have coordinates and respectively and the boundary point where the particle transits from region I to region II have coordinates Note that is fixed and the action depends on the coordinate only. State the expression for the action . Use PLA to obtain the relationship between and these coordinates. The Extremum Principle in Optics II A light ray travels from medium I to medium II with refractive indices and respectively. The two media are separated by a line parallel to the x-axis. The light ray makes an angle with the y-axis in medium I I and in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat’s principle of least time.

Figure 2

A3

1.0

B

B1

The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between and on the basis of Fermat’s principle. Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height.

Figure 3: Tank of Sugar Solution

0.5

B2 B3

Assume that the refractive index depends only on . Use the equation obtained in B1 to obtain the expression for the slope of the beam’s path in terms of refractive index at and . The laser beam is directed horizontally from the origin into the sugar solution at a height from the bottom of the tank as shown in figure 3. Take where and are positive constants. Obtain an expression for in terms of and related quantities for the actual trajectory of the laser beam.

1.5 1.2

1

Manoj Harbola (IIT-Kanpur) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

Q T-2

You may use: ∫

∫√ ( √ ) constant, where ?

Page 2 of 2

or cm,

B4 C

Obtain the value of , the point where the beam meets the bottom of the tank. Take , cm (1 cm = 10-2 m).

0.8

C1

The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from to can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. As the particle moves along its trajectory by an infinitesimal distance , relate the change in the phase of its de Broglie wave to the change in the action and the Planck constant. Recall the problem from part A where the particle traverses from to (see Fig. 4). Let an opaque partition be placed at the boundary AB between the two regions. There is a small opening CD of width in AB such that and . Consider two extreme paths C and D such that C lies on the classical trajectory discussed in part A. Obtain the phase difference between the two paths to first order.

0.6

I D

A

II 1.2

C2

C

B

CD d

II

Figure 4

D

Matter Wave Interference Consider an electron gun at which directs a collimated beam of electrons to a narrow slit at in the opaque partition A B at such that is a straight line. is a point on the screen at (see Fig. 5). The speed in I is × ms and . The potential in II is such that speed × m s . The distance is ( ). Ignore electron-electron interaction.

I

A

215.00 nm

B

Figure 5

250 mm

D1 D2 D3 D4

If the electrons at have been accelerated from rest, calculate the accelerating potential . Another identical slit is made in the partition A B at a distance of nm ( nm m) below slit (Fig. 5). If the phase difference between de Broglie waves arriving at P through the slits F and G is , calculate . What is the smallest distance from P at which null (zero) electron detection maybe expected on the screen? [Note: you may find the approximation useful] The beam has a square cross section of × and the setup is 2 m long. What should be the minimum flux density Imin (number of electrons per unit normal area per unit time) if, on an average, there is at least one electron in the setup at a given time?

0.3 0.8 1.2 0.4

Theoretical Task 2 (T-2) : Solutions

The Extremum Principle1

A. The Extremum Principle in Mechanics Consider a horizontal frictionless x-y plane shown in Fig. 1. It is divided into two regions, I and II, by a line AB satisfying the equation x = x1 . The potential energy of a point particle of mass m in region I is V = 0 while it is V = V0 in region II. The particle is sent from the origin O with speed v1 along a line making an angle θ1 with the x-axis. It reaches point P in region II traveling with speed v2 along a line that makes an angle θ2 with the x?axis. Ignore gravity and relativistic e?ects in this entire task T-2 (all parts). (A1) Obtain an expression for v2 in terms of m, v1 and V0 . Solution: From the principle of Conservation of Mechanical Energy 1 2 1 2 mv = mv + V0 2 1 2 2

2 v2 = (v1 ?

1 of 9

Figure 1 [0.2]

2V0 1/2 ) m [0.3]

(A2) Express v2 in terms of v1 , θ1 and θ2 . Solution: At the boundary there is an impulsive force (∝ dV /dx) in the ?x direction. Hence only the velocity component in the x?direction v1x su?ers change . The component in the y ?direction remains unchanged. Therefore v1y = v2y v1 sin θ1 = v2 sin θ2

We de?ne a quantity called action A = m v (s) ds, where ds is the in?nitesimal length along the trajectory of a particle of mass m moving with speed v (s). The integral is taken over the path. As an example. for a particle moving with constant speed v on a circular path of radius R, the action A for one revolution will be 2πmRv . For a particle with constant energy E , it can be shown that of all the possible trajectories between two ?xed points, the actual trajectory is the one on which A de?ned above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).

Manoj Harbola (IIT-Kanpur) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

1

Theoretical Task 2 (T-2): Solutions

2 of 9

(A3) PLA implies that the trajectory of a particle moving between two ?xed points in a region of constant potential will be a straight line. Let the two ?xed points O and P in Fig. 1 have coordinates (0,0) and (x0 ,y0 ) respectively and the boundary point where the particle transits from region I to region II have coordinates (x1 ,α). Note x1 is ?xed and the action depends on the coordinate α only. State the expression for the action A(α). Use PLA to obtain the the relationship between v1 /v2 and these coordinates. Solution: By de?nition A(α) from O to P is A(α) = mv1

2 x2 1 + α + mv2

[1.0]

(x0 ? x1 )2 + (y0 ? α)2

Di?erentiating w.r.t. α and setting the derivative of A(α) to zero (x2 1 v1 α v2 (y0 ? α) ? =0 2 1 / 2 +α ) [(x0 ? x1 )2 + (y0 ? α)2 ]1/2 ∴

2 1/2 (y0 ? α) (x2 v1 1+α ) = v2 α [(x0 ? x1 )2 + (y0 ? α)2 ]1/2

Note this is the same as A2, namely v1 sin θ1 = v2 sin θ2 .

B. The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices n1 and n2 respectively. The two media are separated by a line parallel to the x-axis. The light ray makes an angle i1 with the y-axis in medium I and i2 in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat’s principle of least time. Figure 2 (B1) The principle states that between two ?xed points, a light ray moves along a path such that the time taken between the two points is an extremum. Derive the relation between sin i1 and sin i2 on the basis of Fermat’s principle. Solution: The speed of light in medium I is c/n1 and in medium II is c/n2 , where c is the speed of light in vacuum. Let the two media be separated by the ?xed line y = y1 . Then time T (α) for light to travel from origin (0,0) and (x0 ,y0 ) is

2 T (α) = n1 ( y1 + α2 )/c + n2 (

[0.5]

(x0 ? α)2 + (y0 ? y1 )2 )/c

Theoretical Task 2 (T-2): Solutions

Di?erentiating w.r.t. α and setting the derivative of T (α) to zero

2 (y1

3 of 9

n1 α n2 (y0 ? α) ? =0 2 1 / 2 +α ) [(x0 ? α)2 + (y0 ? y1 )2 ]1/2 ∴ n1 sin i1 = n2 sin i2

[Note: Derivation is similar to A3. This is Snell’s law.] Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height.

Figure 3

(B2) Assume that the refractive index n(y) depends only on y. Use the equation obtained in B1 to obtain the expresssion for the slope dy/dx of the beam’s path in terms of n0 at y = 0 and n(y ). Solution: From Snell’s law n0 sin i0 = n(y ) sin i dy Then, = ? cot i dx n(y ) 1+( dy =? dx n(y ) n0 sin i0 dy 2 ) dx

2

[1.5]

n0 sin i0 =

?1

(B3) The laser beam is directed horizontally from the origin (0,0) into the sugar solution at a height y0 from the bottom of the tank as shown. Take n(y ) = n0 ? ky where n0 and k are positive constants. Obtain an expression for x in terms of y and related quantities. dx You may use: sec θdθ = ln(sec θ + tan θ) + constant sec θ = 1/ cos θ or √x 2 ?1 = √ 2 ln(x + x ? 1) + constant. Solution: dy =? n0 ? ky 2 ( ) ?1 n0 sin i0 dx

[1.2]

Note i0 = 90o so sin i0 = 1.

Theoretical Task 2 (T-2): Solutions

Method I We employ the substitution n0 ? ky n0 n0 dξ (? ) k =? 2 ξ ?1 ξ= Let ξ = sec θ. Then n0 ln(sec θ + tan θ) = x + c k Or METHOD II We employ the substition n0 ? ky n0 n0 dξ (? ) k =? 2 ξ ?1 ξ= ? n0 ln k n0 ? ky + n0 (

4 of 9

dx

dx

n0 ? ky 2 ) ?1 n0

= ?x + c

Now continuing Considering the substitutions and boundary condition, x = 0 for y = 0 we obtain that the constant c = 0. Hence we obtain the following trajectory: x= n0 ln k n0 ? ky + n0 ( n0 ? ky 2 ) ?1 n0

(B4) Obtain the value of x0 , the point where the beam meets the bottom of the tank. Take y0 = 10.0 cm, n0 = 1.50, k = 0.050 cm?1 (1 cm = 10?2 m). Solution: Given y0 = 10.0 cm. From (B3) x0 = Here y = ?y0 n0 = 1.50 ? k = 0.050 cm?1

1/2

[0.8]

n0 ? n0 ? ky ln k n0

+

n0 ? ky n0

2

? ?

?1

Theoretical Task 2 (T-2): Solutions

n0 (n0 + ky0 ) (n0 + ky0 )2 x0 = ln + ?1 k n0 n2 0 ? ? 1/2 2 2 2 ? = 30 ln ? + ?1 1.5 1.5 4 = 30 ln + 3 = 30 ln 7 9

1/2 1/2

5 of 9

4 + 0.88 3

= 24.0 cm

C. The Extremum Principle and the Wave Nature of Matter We now explore between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from O to P can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. (C1) As the particle moves along its trajectory by an in?nitesimal distance ?s, relate the change ?φ in the phase of its de Broglie wave to the change ?A in the action and the Planck constant. Solution: From the de Broglie hypothesis λ → λdB = h/mv where λ is the de Broglie wavelength and the other symbols have their usual meaning ?φ = = 2π ?s λ

[0.6]

2π mv ?s h 2π ?A = h

Theoretical Task 2 (T-2): Solutions

(C2) Recall the problem from part A where the particle traverses from O to P (see Fig. 4). Let an opaque partition be placed at the boundary AB between the two regions. There is a small opening CD of width d in AB such that d (x0 ? x1 ) and d x1 . Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase di?erence ?φCD between the two paths to ?rst order. Solution:

6 of 9 [1.2]

Figure 4

y

I

A II P D F C B x1 θ2 x

E θ1 O

Consider the extreme trajectories OCP and ODP of (C1) The geometrical path di?erence is ED in region I and CF in region II. This implies (note: d (x0 ? x1 ) and d x1 ) ?φCD = 2πd sin θ1 2πd sin θ2 ? λ1 λ2

?φCD =

2πmv1 d sin θ1 2πmv2 d sin θ2 ? h h md = 2π (v1 sin θ1 ? v2 sin θ2 ) h =0 (from A2 or B1)

Thus near the clasical path there is invariably constructive interference.

Theoretical Task 2 (T-2): Solutions

D. Matter Wave Interference Consider an electron gun at O which directs a collimated beam of electrons to a narrow slit at F in the opaque partition A1 B1 at x = x1 such that OFP is a straight line. P is a point on the screen at x = x0 (see Fig. 5). The speed in I is v1 = 2.0000 × 107 m s?1 and θ = 10.0000? . The potential in region II is such that the speed v2 = 1.9900 × 107 m s?1 . The distance x0 ? x1 is 250.00 mm (1 mm = 10?3 m). Ignore electron-electron interaction.

7 of 9

Figure 5 (D1) If the electrons at O have been accelerated from rest, calculate the accelerating potential U1 . Solution: qU1 = 1 mv 2 2

[0.3]

=

9.11 × 10?31 × 4 × 1014 J 2 = 2 × 9.11 × 10?17 J = 2 × 9.11 × 10?17 eV 1.6 × 10?19 1100 eV )

= 1.139 × 103 eV (

U1 = 1.139 × 103 V (D2) Another identical slit G is made in the partition A1 B1 at a distance of 215.00 nm (1 nm = 10?9 m) below slit F (Fig. 5). If the phase di?erence between de Broglie waves ariving at P from F and G is 2 π β , calculate β .

[0.8]

Solution: Phase di?erence at P is ?φP = 2πd sin θ 2πd sin θ ? λ1 λ2 md sin 10? = 2πβ h

= 2π (v1 ? v2 )

β = 5.13

Theoretical Task 2 (T-2): Solutions

8 of 9

(D3) What is is the smallest distance ?y from P at which null (zero) electron detection maybe expected on the screen? [Note: you may ?nd the approximation sin(θ + ?θ) ≈ sin θ + ?θ cos θ useful] Solution:

y I

F

215 nm

[1.2]

A1 II

P

G

O

B 1 x1

x

From previous part for null (zero) electron detection ?φ = 5.5 × 2π ∴ mv1 d sin θ mv2 d sin(θ + ?θ) ? = 5.5 h h

sin(θ + ?θ) =

= = =

mv1 d sin θ ? 5.5 h mv2 d h h 5.5 v1 sin θ ? v2 m v2 d 5.5 2 sin 10? ? 1.99 1374.78 × 1.99 × 107 × ×2.15 × 10?7 0.174521 ? 0.000935

This yields ?θ = ?0.0036? The closest distance to P is ?y = = = = (x0 ? x1 )(tan(θ + ?θ) ? tan θ) 250(tan 9.9964 ? tan 10) ?0.0162mm ?16.2? m

The negative sign means that the closest minimum is below P. Approximate Calculation for θ and ?y Using the approximation sin(θ + ?θ) ≈ sin θ + ?θ cos θ The phase di?erence of 5.5 × 2π gives mv1 d sin 10? d(sin 10? + ?θ cos 10? ) ? mv2 = 5.5 h h

From solution of the previous part mv1 d sin 10? dsin10? ? mv2 = 5.13 h h

Theoretical Task 2 (T-2): Solutions

Therefore mv2 d?θ cos 10? = 0.3700 h

9 of 9

This yields ?θ ≈ 0.0036? ?y = ?0.0162 mm = ?16.2?m as before

(D4) The electron beam has a square cross section of 500 nm × 500 nm and the setup is 2 m long. What should be the minimum beam ?ux density Imin (number of electrons per unit normal area per unit time) if, on an average, there is at least one electron in the setup at a given time? Solution: The product of the speed of the electrons and number of electron per unit volume on an average yields the intensity. Thus N = 1 = Intensity × Area × Length/ Electron Speed = Imin × 0.25 × 10?12 × 2/2 × 107 This gives Imin = 4× 1019 m?2 s?1 R. Bach, D. Pope, Sy-H Liou and H. Batelaan, New J. of Physics Vol. 15, 033018 (2013).

[0.4]

Q T-3

The Design of a Nuclear Reactor1

235

Page 1 of 2

(Total Marks: 10)

Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being U. Neutron induced fission occurs readily in 235U with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other 235U nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height H and radius R filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural UO2 in solid form of height H, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry.

Fig-I Fig-II Fig-III

Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown).

A

Fuel Pin Data for UO2

1. Molecular weight Mw = 0.270 kg mol-1 3. Melting point Tm = 3.138×103 K

2. Density ρ = 1.060×104 kg m-3 4. Thermal conductivity λ = 3.280 W m-1 K-1

A1

A2 A3

A4

Consider the following fission reaction of a stationary 235U after it absorbs a neutron of negligible kinetic energy. 235 U + 1n → 94Zr + 140Ce + 2 1n + Estimate (in MeV) the total fission energy released. The nuclear masses are: m(235U) = 235.044 u; 94 m( Zr) = 93.9063 u; m(140Ce) = 139.905 u; m(1n) = 1.00867 u and 1 u = 931.502 MeV c-2. Ignore charge imbalance. Estimate N the number of 235U atoms per unit volume in natural UO2. Assume that the neutron flux density, φ = 2.000×1018 m-2 s-1 on the fuel is uniform. The fission crosssection (effective area of the target nucleus) of a 235U nucleus is σf = 5.400×10-26 m2. If 80.00% of the fission energy is available as heat, estimate Q (in W m-3), the rate of heat production in the pin per unit volume. 1MeV = 1.602×10-13 J The steady-state temperature difference between the center (Tc) and the surface (Ts) of the pin can be expressed as Tc?Ts = k F(Q,a,λ), where k = 1 ∕ 4 is a dimensionless constant and a is the radius of the pin. Obtain F(Q,a,λ) by dimensional analysis. Note that λ is the thermal conductivity of UO2.

0.8

0.5 1.2

0.5

1

Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

Q T-3

A5 B

Page 2 of 2

1.0

The desired temperature of the coolant is 5.770×102 K. Estimate the upper limit au on the radius a of the pin. The Moderator Consider the two dimensional elastic collision between a neutron of mass 1 u and a moderator atom of mass A u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let ???? and ???? be the velocities of the neutron before and after collision respectively in the LF. Let ????? be the velocity of the center of mass (CM) frame relative to LF and θ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. The collision in LF is shown schematically, where θL is the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1, 2 and 3 in terms of ???? , ???? and ????? . Indicate the scattering angle θ.

Fig-IV

????

2

B1 ????

1

θL

3

4

Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision

1.0

B2 B3 B4 C

Obtain v and V, the speeds of the neutron and moderator atom in the CM frame after collision, in terms of A and . Derive an expression for G(α, θ) = Ea ∕ Eb , where Eb and Ea are the kinetic energies of the neutron, in the LF, before and after the collision respectively and . Assume that the above expression holds for D2O molecule. Calculate the maximum possible fractional energy loss of the neutron for the D2O (20 u) moderator. The Nuclear Reactor To operate the NR at any constant neutron flux ψ (steady state), the leakage of neutrons has to be compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the leakage rate is k1 [(2.405 ∕ R)2 + (π ∕ H)2] ψ and the excess production rate is k2 ψ. The constants k1 and k2 depend on the material properties of the NR. Consider a NR with k1 = 1.021×10-2 m and k2 = 8.787×10-3 m-1. Noting that for a fixed volume the leakage rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state. The fuel channels are in a square arrangement (Fig-III) with the nearest neighbour distance 0.286 m. The effective radius of a fuel channel (if it were solid) is 3.617×10-2 m. Estimate the number of fuel channels Fn in the reactor and the mass M of UO2 required to operate the NR in steady state.

1.0 1.0 0.5

C1 C2

1.5 1.0

Theoretical Task 3 (T-3) : Solutions

The Design of a Nuclear Reactor1

1 of 9

Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being 235 U. Neutron induced ?ssion occurs readily in 235 U with the emission of 2-3 ?ssion neutrons having high kinetic energy. This ?ssion probability will increase if the neutrons inducing ?ssion have low kinetic energy. So by reducing the kinetic energy of the ?ssion neutrons, one can induce a chain of ?ssions in other 235 U nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height H and radius R ?lled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural UO2 in solid form of height H , are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy, and reach the surrounding fuel channels with low enough energy to cause ?ssion (Figs I-III). Heat generated from ?ssion in the pin is transmitted to a coolant ?uid ?owing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry.

Fig-I

Fig-II

Fig-III

Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown).

A. Fuel Pin Data for UO2 1. Molecular weight Mw =0.270 kg mol 3. Melting point Tm =3.138×103 K

?1

2. Density ρ=1.060×104 kg m?3 4. Thermal conductivity λ=3.280 W m?1 K?1

235

A1 Consider the following ?ssion reaction of a stationary negligible kinetic energy.

235

1

U after it absorbs a neutron of

U +1 n ?→94 Zr +140 Ce + 2 1 n + ?E

Joseph Amal Nathan (BARC) and Vijay A. Singh (ex-National Coordinator, Science Olympiads) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group and the International Board are gratefully acknowledged.

Theoretical Task 3 (T-3) : Solutions

2 of 9

Estimate ?E (in MeV) the total ?ssion energy released. The nuclear masses are: m(235 U) = 235.044 u; m(94 Zr) = 93.9063 u; m(140 Ce) = 139.905 u; m(1 n) = 1.00867 u and 1 u = 931.502 MeV c?2 . Ignore charge imbalance. Solution: ?E = 208.684 MeV Detailed solution: The energy released during the transformation is ?E = [m(235 U) + m(1 n) ? m(94 Zr) ? m(140 Ce) ? 2m(1 n)]c2 Since the data is supplied in terms of uni?ed atomic masses (u), we have ?E = [m(235 U) ? m(94 Zr) ? m(140 Ce) ? m(1 n)]c2

[0.8]

= 208.684 MeV [Acceptable Range (208.000 to 209.000)] from the given data. A2 Estimate N the number of

235

U atoms per unit volume in natural UO2 .

[0.5]

Solution: N = 1.702 × 1026 m?3 Detailed solution: The number of UO2 molecules per m3 of the fuel N1 is given in the terms of its density ρ, the Avogadro number NA and the average molecular weight Mw as ρNA N1 = Mw 10600 × 6.022 × 1023 = 2.364 × 1028 m?3 = 0.270 Each molecule of UO2 contains one uranium atom. Since only 0.72% of these are U, N = 0.0072× N1 = 1.702 × 1026 m?3 [Acceptable Range (1.650 to 1.750)]

235

A3 Assume that the neutron ?ux φ = 2.000 × 1018 m?2 s?1 on the fuel is uniform. The ?ssion cross-section (e?ective area of the target nucleus) of a 235 U nucleus is σf = 5.400 ×10?26 m2 . If 80.00% of the ?ssion energy is available as heat, estimate Q (in W m?3 ) the rate of heat production in the pin per unit volume. 1MeV = 1.602 ×10?13 J. Solution: Q = 4.917 × 108 W/m3 Detailed solution: It is given that 80% of the ?ssion energy is available as heat thus the heat energy available per ?ssion Ef is from a-(i) Ef = 0.8 × 208.7 MeV = 166.96 MeV = 2.675 × 10?11 J The total cross-section per unit volume is N × σf . Thus the heat produced per unit

[1.2]

Theoretical Task 3 (T-3) : Solutions

volume per unit time Q is Q = N × σf × φ × Ef = (1.702 × 1026 ) × (5.4 × 10?26 ) × (2 × 1018 ) × (2.675 × 10?11 ) W/m3 = 4.917 × 108 W/m3 [Acceptable Range (4.800 to 5.000)]

3 of 9

A4 The steady-state temperature di?erence between the center (Tc ) and the surface (Ts ) of the pin can be expressed as Tc ? Ts = kF (Q, a, λ) where k = 1/4 is a dimensionless constant and a is the radius of the pin. Obtain F (Q, a, λ) by dimensional analysis. Solution: Tc ? Ts = Qa2 . 4λ

[0.5]

Detailed solution: The dimensions of Tc ? Ts is temperature. We write this as Tc ? Ts = [K ]. Once can similarly write down the dimensions of Q, a and λ. Equating the temperature to powers of Q, a and λ, one could state the following dimensional equation: K = Qα aβ λγ = [M L?1 T ?3 ] α [L] β [M L1 T ?3 K ?1 ] γ This yields the following algebraic equations γ = -1 equating powers of temperature α + γ = 0 equating powers of mass or time. From the previous equation we get α = 1 Next ?α + β + γ = 0 equating powers of length. This yields β = 2. Qa2 where we insert the dimensionless factor 1/4 as sugThus we obtain Tc ? Ts = 4λ gested in the problem. No penalty if the factor 1/4 is not written. Note: Same credit for alternate ways of obtaining α, β, γ . A5 The desired temperature of the coolant is 5.770 ×102 K. Estimate the upper limit au on the radius a of the pin. Solution: au = 8.267 × 10?3 m. Detailed solution: The melting point of UO2 is 3138 K and the maximum temperature of the coolant is 577 K. This sets a limit on the maximum permissible temperature (Tc ? Ts ) to be less than (3138 - 577 = 2561 K) to avoid “meltdown”. Thus one may take a maximum of (Tc ? Ts ) = 2561 K. Noting that λ = 3.28 W/m - K, we have a2 u = 2561 × 4 × 3.28 4.917 × 108

[1.0]

Where we have used the value of Q from A2. This yields au 8.267 × 10?3 m. So ?3 au = 8.267 × 10 m constitutes an upper limit on the radius of the fuel pin. Note: The Tarapur 3 & 4 NR in Western India has a fuel pin radius of 6.090 × 10?3 m.

Theoretical Task 3 (T-3) : Solutions

4 of 9

B. The Moderator Consider the two dimensional elastic collision between a neutron of mass 1 u and a moderator atom of mass A u. Before collision all the moderator atoms are considered at rest in the ? ? laboratory frame (LF). Let → vb and → va be the velocities of the neutron before and after collision ? → respectively in the LF. Let vm be the velocity of the center of mass (CM) frame relative to LF and θ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at non-relativistic speeds B1 The collision in LF is shown schematically with θL as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1, 2 and 3 in ? ? →. Indicate the scattering angle θ. terms of → vb , → va and ? v m

Fig-IV

[1.0]

? va

2

? vb

1

?L

3

4

Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision

Solution:

Laboratory Frame

Center of Mass Frame

? va

2

? ? va ? vm ? ? vb ? vm

? vb

1

?L

3

4

?

? ? vm

B2 Obtain v and V , the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of A and vb . Solution: Detailed solution: Before the collision in the CM frame (vb ? vm ) and vm will be magnitude of the velocities of the neutron and moderator atom respectively. b From momentum conservation in the CM frame, vb ? vm = Avm gives vm = Av . +1 After the collision, let v and V be magnitude of the velocities of neutron and moderator atom respectively in the CM frame. From conservation laws, v = AV and 1 1 2 1 1 (vb ? vm )2 + Avm = v 2 + AV 2 .(→ [0.2 + 0.2]) 2 2 2 2

[1.0]

Theoretical Task 3 (T-3) : Solutions

5 of 9

Avb b and V = Av . (OR) From de?nition of center of mass frame Solving gives v = A +1 +1 vb Avb vm = A+1 . Before the collision in the CM frame vb ? vm = A and vm will be mag+1 nitude of the velocities of the neutron and moderator atom respectively. In elastic collision the particles are scattered in the opposite direction in the CM frame and so Avb b and V = Av (→ [0.2 + 0.1]). the speeds remain same v = A +1 +1

Note: Alternative solutions are worked out in the end and will get appropriate weightage. B3 Derive an expression for G(α, θ) = Ea /Eb , where Eb and Ea are the kinetic energies of the neutron, in the LF, before and after the collision respectively, and α ≡ [(A ? 1)/(A + 1)]2 , Solution: G(α, θ) = 1 A2 + 2A cos θ + 1 Ea = [(1 + α) + (1 ? α) cos θ] . = 2 Eb (A + 1) 2

[1.0]

? ? →, v 2 = v 2 + v 2 + 2vv cos θ (→ [0.3]). SubstiDetailed solution: Since → va = → v +? v m m a m 2 2 2 2 A vb vb 2Avb 2 tuting the values of v and vm , va = (A+1)2 + (A+1)2 + (A+1)2 cos θ (→ [0.2]), so

2 A2 + 2A cos θ + 1 va Ea = . = 2 vb Eb (A + 1)2

G(α, θ) = Alternate form

A2 + 1 2A 1 + cos θ = [(1 + α) + (1 ? α) cos θ] . 2 2 (A + 1) (A + 1) 2 (1 ? α)(1 ? cos θ) . 2

= 1?

Note: Alternative solutions are worked out in the end and will get appropriate weightage. B4 Assume that the above expression holds for D2 O molecule. Calculate the maximum pos?Ea sible fractional energy loss fl ≡ EbE of the neutron for the D2 O (20 u) moderator. b Solution: fl = 0.181 Detailed solution: The maximum energy loss will be when the collision is head on ie., Ea will be minimum for the scattering angle θ = π . So Ea = Emin = αEb . For D2 O, α = 0.819 and maximum fractional loss ceptable Range (0.170 to 0.190)]

Eb ?Emin Eb

[0.5]

= 1 ? α = 0.181. [Ac-

Theoretical Task 3 (T-3) : Solutions

6 of 9

C. The Nuclear Reactor To operate the NR at any constant neutron ?ux Ψ (steady state), the leakage of neutrons has to be compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical 405 2 π 2 geometry the leakage rate is k1 2.R + H Ψ and the excess production rate is k2 Ψ. The constants k1 and k2 depend on the material properties of the NR. C1 Consider a NR with k1 = 1.021×10?2 m and k2 = 8.787×10?3 m?1 . Noting that for a ?xed volume the leakage rate is to be minimized for e?cient fuel utilisation obtain the dimensions of the NR in the steady state. Solution: R = 3.175 m, H = 5.866 m. Detailed solution: For constant volume V = πR2 H , d dH 2.405 R

2

[1.5]

+

π H

2

= 0,

π2 π2 d 2.4052 πH 2.4052 π + 2 = ? 2 3 = 0, dH V H V H

2.405 2 R π 2 . H

gives

=2

For steady state, 1.021 × 10?2 2.405 R

2

+

π H

2

Ψ = 8.787 × 10?3 Ψ.

Hence H = 5.866 m [Acceptable Range (5.870 to 5.890)] R = 3.175 m [Acceptable Range (3.170 to 3.180)].

Alternative Non-Calculus Method to Optimize 2 2.405 π 2 Minimisation of the expression + , for a ?xed volume V = R H πR2 H : 2.4052 πH π2 Substituting for R2 in terms of V, H we get + 2, V H 2 2 2 2.405 πH 2.405 πH π which can be written as, + + 2. 2V 2V H Since all the terms are positive applying AMGM inequality for three positive terms we get

2.4052 πH 2V

+

2.4052 πH 2V

+

π2 H2

3

≥

3

2.4052 πH 2.4052 πH π2 × × 2 = 2V 2V H

3

2.4054 π 4 . 4V 2

Theoretical Task 3 (T-3) : Solutions

7 of 9

The RHS is a constant. The LHS is always greater or equal to this constant implies that this is the minimum value the LHS can achieve. The minimum is achieved 2.4052 πH = when all the three positive terms are equal, which gives the condition 2V 2 π2 2.405 π 2 ? = 2 . H2 R H For steady state, 1.021 × 10?2 2.405 R

2

+

π H

2

Ψ = 8.787 × 10?3 Ψ.

Hence H = 5.866 m [Acceptable Range (5.870 to 5.890)] R = 3.175 m [Acceptable Range (3.170 to 3.180)]. π2 . From the condiH2

Note: Putting the condition in the RHS gives the minimum as tion we get π3 2.4052 π 2 π2 = ? = H3 2V H2

3

2.4054 π 4 . 4V 2

Note: The radius and height of the Tarapur 3 & 4 NR in Western India is 3.192 m and 5.940 m respectively. C2 The fuel channels are in a square arrangement (Fig-III) with nearest neighbour distance 0.286 m. The e?ective radius of a fuel channel (if it were solid) is 3.617 × 10?2 m. Estimate the number of fuel channels Fn in the reactor and the mass M of UO2 required to operate the NR in steady state. Solution: Fn = 387 and M = 9.892 × 104 kg. Detailed solution: Since the fuel channels are in square pitch of 0.286 m, the effective area per channel is 0.2862 m2 = 8.180 × 10?2 m2 . The cross-sectional area of the core is πR2 = 3.142 × (3.175)2 = 31.67 m2 , so the maximum number of fuel channels that can be accommodated in the cylinder is the .67 integer part of 031 = 387. .0818 Mass of the fuel=387×Volume of the rod×density = 387 × (π × 0.036172 × 5.866) × 10600 = 9.892 × 104 kg. Fn = 387 [Acceptable Range (380 to 394)] M = 9.892 × 104 kg [Acceptable Range (9.000 to 10.00)]

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Note 1: (Not part of grading) The total volume of the fuel is 387 × (π × 0.036172 × 5.866) = 9.332 m3 . If the reactor works at 12.5 % e?cieny then using the result of a-(iii) we have that the power output of the reactor is 9.332 × 4.917 × 108 × 0.125 =

Theoretical Task 3 (T-3) : Solutions

573 MW.

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Note 2: The Tarapur 3 & 4 NR in Western India has 392 channels and the mass of the fuel in it is 10.15 ×104 kg. It produces 540 MW of power. Alternative Solutions to sub-parts B2 and B3: Let σ be the scattering angle of the Moderator atom in the LF, taken clockwise with respect to the initial direction of the neutron before collision. Let U be the speed of the Moderator atom, in the LF, after collision. From momentum and kinetic conservation in LF we have vb = va cos θL + AU cos σ, 0 = va sin θL ? AU sin σ, 1 1 2 1 2 . vb = AU 2 + va 2 2 2 Squaring and adding eq(1) and (2) to eliminate σ and from eq(3) we get

2 2 A 2 U 2 = va + vb ? 2va vb cos θL , 2 2 2 2 A U = Avb ? Ava ,

(1) (2) (3)

(4) (5)

which gives

2 2 2va vb cos θL = (A + 1)va ? (A ? 1)vb .

(ii) Let v be the speed of the neutron after collision in the COMF. From de?nition of center vb of mass frame vm = . A+1 va sin θL and va cos θL are the perpendicular and parallel components of va , in the LF, resolved along the initial direction of the neutron before collision. Transforming these to the COMF gives va sin θL and va cos θL ? vm as the perpendicular and parallel components of v . Substitut2 sin2 θ + v 2 cos2 θ + v 2 ? 2v v cos θ ing for vm and for 2va vb cos θL from eq(5) in v = va L L a m L a m Avb 2 . Squaring the components of v to eliminate θL gives va = and simplifying gives v = A+1 2 v 2 + vm + 2vvm cos θ. Substituting for v and vm and simplifying gives,

2 va Ea A2 + 2A cos θ + 1 . = = 2 vb Eb (A + 1)2

Ea A2 + 1 2A 1 G(α, θ) = = + cos θ = [(1 + α) + (1 ? α) cos θ] . 2 2 Eb (A + 1) (A + 1) 2 (OR) vb . After the collision, let v and V A+1 be magnitude of the velocities of neutron and moderator atom respectively in the COMF. From conservation laws in the COMF, (iii) From de?nition of center of mass frame vm = v = AV and 1 1 1 2 1 (vb ? vm )2 + Avm = v 2 + AV 2 . 2 2 2 2

Avb b Solving gives v = A and V = Av . We also have v cos θ = va cos θL ? vm , substituting for vm +1 +1 and for va cos θL from eq(5) and simplifying gives 2 va Ea A2 + 2A cos θ + 1 = = . 2 vb Eb (A + 1)2

Theoretical Task 3 (T-3) : Solutions

G(α, θ) = (OR) (iv) From de?nition of center of mass frame vm = A2 + 1 Ea 2A 1 = + cos θ = [(1 + α) + (1 ? α) cos θ] . 2 2 Eb (A + 1) (A + 1) 2

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vb . After the collision, let v and V A+1 be magnitude of the velocities of neutron and moderator atom respectively in the CM frame. From conservation laws in the CM frame, v = AV and 1 1 2 1 1 (vb ? vm )2 + Avm = v 2 + AV 2 . 2 2 2 2

Avb b Solving gives v = A and V = Av . U sin σ and U cos σ are the perpendicular and parallel +1 +1 components of U , in the LF, resolved along the initial direction of the neutron before collision. Transforming these to the COMF gives U sin σ and ?U cos σ + vm as the perpendicular and 2 ? 2V vm cos θ. Since V = vm parallel components of V . So we get U 2 = V 2 sin2 θ + V 2 cos2 θ + vm 2 2 we get U = 2vm (1 ? cos θ). Substituting for U from eq(4) and simplifying gives 2 va Ea A2 + 2A cos θ + 1 = = . 2 vb Eb (A + 1)2

G(α, θ) =

A2 + 1 Ea 2A 1 = + cos θ = [(1 + α) + (1 ? α) cos θ] . 2 2 Eb (A + 1) (A + 1) 2

√ A2 + 2A cos θ + 1 Note: We have va = vb . Substituting for va , v, vm in v cos θ = va cos θL ?vm A+1 gives the relation between θL and θ, cos θL = √ A2 A cos θ + 1 . + 2A cos θ + 1

Treating the above equation as quadratic in cos θ gives, cos θ = ? sin2 θL ± cos θL A A2 ? sin2 θL .

For θL = 0? the root with the negative sign gives θ = 180? which is not correct so, cos θ = cos θL A2 ? sin2 θL ? sin2 θL . A

2 va gives an expression in terms 2 vb

Substituting the above expression for cos θ in the expression for of cos θL

2 va Ea A2 + 2 cos θL A2 ? sin2 θL + cos 2θL = . = 2 vb Eb (A + 1)2