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Theoretical Task 1 (T-1) : Solutions
1 of 7
Particles from the Sun1
Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures
and also confirm 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:
�
x
1
− 2 eax + constant
a a
�
� 2
�
x
2x
2
− 2 + 3 eax + constant
(ii)
x2 eax dx =
a
a
a
�
� 3
�
2
x
6x
6
3x
− 2 + 3 − 4 eax + constant
(iii)
x3 eax dx =
a
a
a
a
(i)
�
xeax dx =
�
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.
[0.3]
Solution:
2
)(σTs4 )
Stefan’s law: L� = (4πR�
Ts =
�
L�
2
4πR�
σ
�1/4
= 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
u(ν) = A 2� 2 ν 3 exp(−hν/kB Ts ),
d� c
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 .
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.
!!;
[0.3]
!"#$%&'()*%)+(&
Theoretical Task 1 (T-1) : Solutions
Solution:
Pin =
�
∞
u(ν)dν =
0
�
∞
A
0
2 of 7
2
R�
2πh 3
ν exp(−hν/kB Ts )dν
d2� c2
kB Ts
kB T s
hν
Let x =
x
dν =
dx.
. Then, ν =
kB T s
h
h
�
2
2πhAR�
R2
R2
(kB Ts )4 ∞ 3 −x
2πk 4
12πk 4
Pin =
x e dx = 2 3B Ts 4 A 2� · 6 = 2 3B Ts 4 A 2�
c2 d2�
h4
c
h
d
c
h
d�
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 .
[0.2]
Solution:
u(ν)
hν
R2 2π
= A 2� 2 ν 2 exp(−hν/kB Ts )
d� c
nγ (ν) =
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) Define 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 .
[1.0]
Solution:
The useful power output is the useful energy quantum per photon, Eg ≡ hνg , multiplied by
the number of photons with energy, E ≥ Eg .
� ∞
Pout = hνg
nγ (ν)dν
νg
= hνg A
2
R�
2π
d2� c2
= kB Ts xg A
=
�
∞
νg
2
R�
2π
2
d � c2
ν 2 exp(−hν/kB Ts )dν
�
kB T s
h
�3 �
∞
x2 e−x dx
xg
2
2πkB4 4 R�
Ts A 2 xg (x2g + 2xg + 2)e−xg
c2 h 3
d�
(A5) Express the efficiency, η, of this solar cell in terms of xg .
!!!
[0.2]
,-./%)+(&*%0"1234*%,5643
Theoretical Task 1 (T-1) : Solutions
3 of 7
Solution:
Efficiency η =
Pout
xg 2
(x + 2xg + 2)e−xg
=
Pin
6 g
(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 → ∞
[1.0]
Solution:
1 3
(x + 2x2g + 2xg )e−xg
6 g
Put limiting values, η(0) = 0
η(∞) = 0.
Since the polynomial has all positive coefficients, it increases monotonically; the exponential
function decreases monotonically. Therefore, η has only one maximum.
η=
η
dη
1
= (−x3g + x2g + 2xg + 2)e−xg
dxg
6
�
�
dη ��
dη ��
1
=
=0
�
dxg xg =0 3
dxg �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
1
dη
= (−x3g + x2g + 2xg + 2)e−xg = 0
dxg
6
⇒ p(xg ) ≡ x3g − x2g − 2xg − 2 = 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.
!!"
[1.0]
!"#$%&'()*%)+(&
Theoretical Task 1 (T-1) : Solutions
4 of 7
Alternative methods leading to the same result are acceptable.
η(2.27) = 0.457
(A8) The band gap of pure silicon is Eg = 1.11 eV. Calculate the efficiency, ηSi , of a silicon solar cell
using this value.
Solution:
[0.2]
1.11 × 1.60 × 10−19
= 2.23
1.38 × 10−23 × 5763
xg
ηSi = (x2g + 2xg + 2)e−xg = 0.457
6
xg =
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, Ω, of the Sun at present, in terms of G, M� and R� .
Solution:
The total gravitational potential energy of the Sun: Ω = −
For constant density, ρ =
Ω=−
�
R�
0
G
3M�
3
4πR�
�
4
m = πr3 ρ
3
�
M�
0
[0.3]
Gm dm
r
dm = 4πr2 ρdr
�
5
� dr
16π 2 Gρ2 R�
3 GM�2
4 3 �
πr ρ 4πr2 ρ
=−
=−
3
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 =
−Ω
L�
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.
!!H
[0.5]
,-./%)+(&*%0"1234*%,5643
Theoretical Task 1 (T-1) : Solutions
5 of 7
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:
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 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, 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:
⇒ Φν =
[0.6]
4.0 × 10−12 J ↔ 2ν
3.85 × 1026
L�
×2=
× 2 = 6.8 × 1014 m−2 s−1 .
4πd2� δE
4π × (1.50 × 1011 )2 × 4.0 × 10−12
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 efficiency of the detector for detecting νx is 1/6th of its efficiency
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:
N1
Ne
Nx
N2
=
=
=
=
�N0
�N0 (1 − f )
�N0 f /6
Ne + Nx
OR
f
(1 − f )N1 + N1 = N2
6
�
�
N2
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Theoretical Task 1 (T-1) : Solutions
6 of 7
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, 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.)
[2.0]
Solution:
When the electron stops emitting Cherenkov radiation, its speed has reduced to vstop = c/n.
Its total energy at this time is
me c 2
nme c2
=√
Estop = �
2
n2 − 1
1 − vstop
/c2
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
n2 − 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 fluctuates 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]
!!J
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Theoretical Task 1 (T-1) : Solutions
7 of 7
Solution:
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.
1
3
2
mBe VBe
= kB Tc
2
2
⇒
Tc = 1.13 × 107 K
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Theoretical Task 2 (T-2) : Solutions
1 of 9
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 effects in this entire task
T-2 (all parts).
Figure 1
(A1) Obtain an expression for v2 in terms of m, v1 and V0 .
[0.2]
Solution:
From the principle of Conservation of Mechanical Energy
1 2 1 2
mv = mv + V0
2 1 2 2
v2 = (v12 −
2V0 1/2
)
m
(A2) Express v2 in terms of v1 , θ1 and θ2 .
[0.3]
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 suffers change . The component in
the y−direction remains unchanged. Therefore
v1y = v2y
v1 sin θ1 = v2 sin θ2
�
We define a quantity called action A = m v(s) ds, where ds is the infinitesimal 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 fixed points, the actual trajectory is
the one on which A defined above is an extremum (minimum or maximum). Historically this
is known as the Principle of Least Action (PLA).
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.
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Theoretical Task 2 (T-2): Solutions
2 of 9
(A3) 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 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 fixed 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.
[1.0]
Solution:
By definition A(α) from O to P is
�
�
A(α) = mv1 x21 + α2 + mv2 (x0 − x1 )2 + (y0 − α)2
Differentiating w.r.t. α and setting the derivative of A(α) to zero
v1 α
v2 (y0 − α)
−
=0
(x21 + α2 )1/2 [(x0 − x1 )2 + (y0 − α)2 ]1/2
∴
(y0 − α) (x21 + α2 )1/2
v1
=
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 fixed 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 fixed line y = y1 . Then time T (α) for light to travel from origin
(0,0) and (x0 ,y0 ) is
�
�
T (α) = n1 ( y12 + α2 )/c + n2 ( (x0 − α)2 + (y0 − y1 )2 )/c
!";
[0.5]
!"#$%&'()*%)+(&
Theoretical Task 2 (T-2): Solutions
3 of 9
Differentiating w.r.t. α and setting the derivative of T (α) to zero
n1 α
n2 (y0 − α)
−
=0
(y12 + α2 )1/2 [(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).
[1.5]
Solution:
From Snell’s law n0 sin i0 = n(y) sin i
dy
Then,
= − cot i
dx
n0 sin i0 = �
n(y)
dy 2
)
dx
��
�2
n(y)
dy
=−
−1
dx
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.
�
You may use:
sec θdθ = ln(sec θ + tan θ) + constant sec θ = 1/ cos θ or √xdx
2 −1 =
√
ln(x + x2 − 1) + constant.
Solution:
�
dy
�
=−
n0 − ky 2
(
) −1
n0 sin i0
Note i0 = 90o so sin i0 = 1.
!"!
�
dx
[1.2]
,-./%)+(&*%0"1234*%,5643
Theoretical Task 2 (T-2): Solutions
4 of 9
Method I We employ the substitution
n0 − ky
n0
� dξ(− n0 )
�
� k = − dx
ξ2 − 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
n
0
� dξ(− )
�
� k = − dx
2
ξ −1
ξ=
−
n0
ln
k
�
n0 − ky
+
n0
�
(
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:
�
�
�
n0 − ky
n0 − ky 2
n0
ln
+ (
) −1
x=
k
n0
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)
Here y = −y0
n0 = 1.50
k = 0.050 cm−1

�1/2 
�
� ��
�2
n0 − ky
n0  n0 − ky

ln
+
−1
x0 =
k
n0
n0
!""
[0.8]
!"#$%&'()*%)+(&
Theoretical Task 2 (T-2): Solutions
x0 =
5 of 9
�
�
�1/2 �
(n0 + ky0 )
(n0 + ky0 )2
n0
ln
+
−
1
k
n0
n20

�� �
�1/2 
2
2
2

= 30 ln 
+
−1
1.5
1.5
�
� �1/2 �
7
4
+
= 30 ln
3
9
�
�
4
+ 0.88
= 30 ln
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 infinitesimal 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
2π
mv∆s
h
2π∆A
=
h
=
!"H
[0.6]
,-./%)+(&*%0"1234*%,5643
Theoretical Task 2 (T-2): Solutions
(C2)
[1.2]
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 difference ∆φCD between
the two paths to first order.
Figure 4
Solution:
y
A II
I
P
E
D
F
C
B
θ1
θ2
x1
O
x
Consider the extreme trajectories OCP and ODP of (C1)
The geometrical path difference 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
2πmv1 d sin θ1 2πmv2 d sin θ2
−
h
h
md
(v1 sin θ1 − v2 sin θ2 )
= 2π
h
=0
(from A2 or B1)
∆φCD =
Thus near the clasical path there is invariably constructive interference.
!"L
6 of 9
!"#$%&'()*%)+(&
Theoretical Task 2 (T-2): Solutions
7 of 9
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.
Figure 5
(D1) If the electrons at O have been accelerated from rest, calculate the accelerating potential
U1 .
Solution:
qU1 =
=
[0.3]
1
mv 2
2
9.11 × 10−31 × 4 × 1014
J
2
−17
= 2 × 9.11 × 10 J
=
2 × 9.11 × 10−17
eV
1.6 × 10−19
= 1.139 × 103 eV (� 1100 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 difference between de Broglie waves ariving
at P from F and G is 2 π β, calculate β.
Solution: Phase difference at P is
∆φP =
2πd sin θ 2πd sin θ
−
λ1
λ2
= 2π(v1 − v2 )
md
sin 10◦ = 2πβ
h
β = 5.13
!"J
[0.8]
,-./%)+(&*%0"1234*%,5643
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 find the approximation sin(θ + ∆θ) ≈ sin θ +
∆θ cos θ useful]
Solution:
I
y
A1 II
P
F
215 nm
G
B1
x1
O
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
mv1 d sin θ
− 5.5
h
mv2 d
h
v1
h 5.5
=
sin θ −
v2
m v2 d
2
5.5
=
sin 10◦ −
1.99
1374.78 × 1.99 × 107 × ×2.15 × 10−7
= 0.174521 − 0.000935
sin(θ + ∆θ) =
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
−16.2µ 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 difference 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
!"<
[1.2]
!"#$%&'()*%)+(&
Theoretical Task 2 (T-2): Solutions
Therefore
mv2
9 of 9
d∆θ cos 10◦
= 0.3700
h
This yields ∆θ ≈ 0.0036◦
∆y = −0.0162 mm = −16.2µm 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 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?
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).
!"R
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Theoretical Task 3 (T-3) : Solutions
1 of 9
The Design of a Nuclear Reactor1
Uranium occurs in nature as UO2 with only 0.720% of the uranium atoms being 235 U. Neutron
induced fission occurs readily in 235 U 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 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 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.
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A. Fuel Pin
Data for UO2
1. Molecular weight Mw =0.270 kg mol−1
2. Density ρ=1.060×104 kg m−3
3. Melting point Tm =3.138×103 K
4. Thermal conductivity λ=3.280 W m−1 K−1
A1 Consider the following fission reaction of a stationary
negligible kinetic energy.
235
1
235
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.
!H;
!"#$%&'()*%)+(&
Theoretical Task 3 (T-3) : Solutions
2 of 9
Estimate ∆E (in MeV) the total fission 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.
[0.8]
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 unified atomic masses (u), we have
∆E = [m(235 U) − m(94 Zr) − m(140 Ce) − m(1 n)]c2
= 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
235
U,
N = 0.0072× N1
= 1.702 × 1026 m−3 [Acceptable Range (1.650 to 1.750)]
A3 Assume that the neutron flux φ = 2.000 × 1018 m−2 s−1 on the fuel is uniform. The fission
cross-section (effective area of the target nucleus) of a 235 U 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.
Solution: Q = 4.917 × 108 W/m3
Detailed solution: It is given that 80% of the fission energy is available as heat
thus the heat energy available per fission 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
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[1.2]
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Theoretical Task 3 (T-3) : Solutions
3 of 9
volume per unit time Q is
Q = N × σ f × 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)]
A4 The steady-state temperature difference 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 =
[0.5]
Qa2
.
4λ
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
Thus we obtain Tc − Ts =
where we insert the dimensionless factor 1/4 as sug4λ
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
a2u =
2561 × 4 × 3.28
4.917 × 108
Where we have used the value of Q from A2. This yields au � 8.267 × 10−3 m. So
au = 8.267 × 10−3 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.
!H"
[1.0]
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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 −
v→
m 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
→
→
terms of −
vb , −
va and −
v→
m . Indicate the scattering angle θ.
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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.
vb
From momentum conservation in the CM frame, vb − vm = Avm gives vm = A+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 2
1
1
1
(vb − vm )2 + Avm
= v 2 + AV 2 .(→ [0.2 + 0.2])
2
2
2
2
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[1.0]
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Theoretical Task 3 (T-3) : Solutions
5 of 9
Avb
vb
Solving gives v = A+1
and V = A+1
. (OR) From definition of center of mass frame
vb
Avb
. Before the collision in the CM frame vb − vm = A+1
and vm will be magvm = A+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
vb
the speeds remain same v = A+1
and V = A+1
(→ [0.2 + 0.1]).
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 ,
[1.0]
Solution:
G(α, θ) =
Ea
A2 + 2A cos θ + 1
1
=
= [(1 + α) + (1 − α) cos θ] .
Eb
(A + 1)2
2
→
→
2
2
2
va = −
v +−
v→
Detailed solution: Since −
m , va = v + vm + 2vvm cos θ (→ [0.3]). Substivb2
2Av 2
A2 vb2
2
tuting the values of v and vm , va = (A+1)2 + (A+1)2 + (A+1)b 2 cos θ (→ [0.2]), so
va2
Ea
A2 + 2A cos θ + 1
=
=
.
2
vb
Eb
(A + 1)2
G(α, θ) =
A2 + 1
2A
1
+
cos θ = [(1 + α) + (1 − α) cos θ] .
(A + 1)2 (A + 1)2
2
Alternate form
= 1−
(1 − α)(1 − cos θ)
.
2
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 posa
sible fractional energy loss fl ≡ EbE−E
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)]
!HL
�
Eb −Emin
Eb
�
= 1 − α = 0.181. [Ac-
[0.5]
!"#$%&'()*%)+(&
Theoretical Task 3 (T-3) : Solutions
6 of 9
C. 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
��
�2 � π �2 �
geometry the leakage rate is k1 2.405
+ H
Ψ and the excess production rate is k2 Ψ. The
R
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
fixed volume the leakage rate is to be minimized for efficient 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,
��
�2 � � �
2.405
π 2
d
= 0,
+
dH
R
H
�
�
d 2.4052 πH
π2
π2
2.4052 π
+ 2 =
− 2 3 = 0,
dH
V
H
V
H
gives
� 2.405 �2
R
=2
For steady state,
� π �2
H
.
1.021 × 10−2
��
2.405
R
�2
+
� π �2
H
�
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
�2 � �
� to Optimize
π 2
2.405
, for a fixed volume V =
+
Minimisation of the expression
R
H
πR2 H:
π2
2.4052 πH
Substituting for R2 in terms of V, H we get
+ 2,
V
H
π2
2.4052 πH 2.4052 πH
+
+ 2.
which can be written as,
2V
2V
H
Since all the terms are positive applying AMGM inequality for three positive terms we
get
�
�
2 πH
2
2.4052 πH
2
4 4
+ 2.405
+ Hπ 2
2.4052 πH
π2
3 2.405 πH
3 2.405 π
2V
2V
≥
×
× 2 =
.
3
2V
2V
H
4V 2
!HJ
[1.5]
,-./%)+(&*%0"1234*%,5643
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
�
� π �2
π2
2.405
=2
.
⇒
H2
R
H
For steady state,
1.021 × 10−2
��
2.405
R
�2
+
� π �2
H
�
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
Note: Putting the condition in the RHS gives the minimum as 2 . From the condiH
�
4 4
π2
2.4052 π 2
π3
3 2.405 π
⇒ 2 =
.
tion we get 3 =
H
2V
H
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 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.
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
31.67
integer part of 0.0818
= 387.
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)]
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 % efficieny then using the result of
a-(iii) we have that the power output of the reactor is 9.332 × 4.917 × 108 × 0.125 =
!H<
[1.0]
!"#$%&'()*%)+(&
Theoretical Task 3 (T-3) : Solutions
8 of 9
573 MW.
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
1 2
vb =
AU 2 + va2 .
2
2
2
(1)
(2)
(3)
Squaring and adding eq(1) and (2) to eliminate σ and from eq(3) we get
A2 U 2 = va2 + vb2 − 2va vb cos θL ,
A2 U 2 = Avb2 − Ava2 ,
(4)
2va vb cos θL = (A + 1)va2 − (A − 1)vb2 .
(5)
which gives
(ii) Let v be the speed of the neutron after collision in the COMF. From definition 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 − 2v v cos θ
ing for vm and for 2va vb cos θL from eq(5) in v = va2 sin2 θL + va2 cos2 θL + vm
a m
L
Avb
. Squaring the components of v to eliminate θL gives va2 =
and simplifying gives v =
A+1
2
v 2 + vm
+ 2vvm cos θ. Substituting for v and vm and simplifying gives,
va2
Ea
A2 + 2A cos θ + 1
=
=
.
vb2
Eb
(A + 1)2
G(α, θ) =
A2 + 1
2A
1
Ea
=
+
cos θ = [(1 + α) + (1 − α) cos θ] .
Eb
(A + 1)2 (A + 1)2
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 definition of center of mass frame vm =
v = AV
and
1
1 2
1
1
(vb − vm )2 + Avm
= v 2 + AV 2 .
2
2
2
2
Avb
vb
Solving gives v = A+1
and V = A+1
. We also have v cos θ = va cos θL − vm , substituting for vm
and for va cos θL from eq(5) and simplifying gives
va2
Ea
A2 + 2A cos θ + 1
=
=
.
2
vb
Eb
(A + 1)2
!HR
,-./%)+(&*%0"1234*%,5643
Theoretical Task 3 (T-3) : Solutions
G(α, θ) =
9 of 9
Ea
A2 + 1
2A
1
=
+
cos θ = [(1 + α) + (1 − α) cos θ] .
Eb
(A + 1)2 (A + 1)2
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 CM frame.
From conservation laws in the CM frame,
(iv) From definition of center of mass frame vm =
v = AV
and
1
1 2
1
1
(vb − vm )2 + Avm
= v 2 + AV 2 .
2
2
2
2
Avb
vb
Solving gives v = A+1
and V = A+1
. U sin σ and U cos σ are the perpendicular and parallel
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
parallel components of V . So we get U 2 = V 2 sin2 θ +V 2 cos2 θ +vm
−2V vm cos θ. Since V = vm
2
2
we get U = 2vm (1 − cos θ). Substituting for U from eq(4) and simplifying gives
va2
Ea
A2 + 2A cos θ + 1
=
=
.
2
vb
Eb
(A + 1)2
G(α, θ) =
A2 + 1
2A
1
Ea
=
+
cos θ = [(1 + α) + (1 − α) cos θ] .
Eb
(A + 1)2 (A + 1)2
2
√
A2 + 2A cos θ + 1
vb . Substituting for va , v, vm in v cos θ = va cos θL −vm
A+1
gives the relation between θL and θ,
Note: We have va =
cos θL = √
A cos θ + 1
.
A2 + 2A cos θ + 1
Treating the above equation as quadratic in cos θ gives,
�
− sin2 θL ± cos θL A2 − sin2 θL
.
cos θ =
A
For θL = 0◦ the root with the negative sign gives θ = 180◦ which is not correct so,
�
cos θL A2 − sin2 θL − sin2 θL
.
cos θ =
A
Substituting the above expression for cos θ in the expression for
of cos θL
va2
gives an expression in terms
vb2
�
va2
Ea
A2 + 2 cos θL A2 − sin2 θL + cos 2θL
=
=
.
vb2
Eb
(A + 1)2
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JULY 05, 2015
ISSUE
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JULY 06, 2015
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ISSUE
ISSUE
RVW RI \RX PLJKW NQRZ ZKDW WR H[SHFW GXULQJ \RXU KHULWDJH ZDON 6RXWK 0XPEDL RIIHUV D
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GHGLFDWHG WR EDVLF UHVHDUFK LQ PDWKHPDWLFV DQG WKH VFLHQFHV ,W LV DOVR D 'HHPHG 8QLYHUVLW\ DQG
ZRUNV XQGHU WKH XPEUHOOD RI WKH 'HSDUWPHQW RI $WRPLF (QHUJ\ RI WKH *RYHUQPHQW RI ,QGLD 7KH
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ISSUE
!
JULY 12, 2015
ISSUE
!
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