Bangladesh Physics Olympiad 2016

SCIENCE OLYMPIAD BLOG

2016

BANGLADESH PHYSICS OLYMPIAD PROBLEMS

6th National Contest, BdPhO C category Time-3 hours 1st Jan, 2016

Instructions Make sure your name and other identification tags are written clearly on the script. Also make sure that the answers to the different parts of the questions below are clearly labelled, otherwise BdPhO will not be held responsible for the consequences.

The Exam- 25 marks total 1. Electron on an elliptic track - 10 marks : The Bohr model for Hydrogen atoms assumes the electron orbits to be circular. This is why it fails to explain the observed fine structure of the atomic spectra hydrogen which was later explained by Arnold Sommerfeld. Sommerfeld considered elliptic orbits for electrons and also took account of relativistic effects. Interestingly, this is an incorrect explanation which was realized only after the development of quantum mechanics. The success of Sommerfeld’s work was nothing but an amazing coincidence. In Sommerfeld’s picture the nucleus is assumed to be placed at one of the focus of the elliptical orbits followed by the electrons. The different elliptical orbits are classified by another quantum number k , which was called the azimuthal quantum number -which is similar to the principal quantum number n related to the energy of the orbit. The force between the nucleus and the electron is caused by electrostatic attraction as in Bohr’s model. To facilitate our calculations below, we recommend that one sets the proportionality 1 constant in Coulomb force law k to be equal to one ( in other words we replace 4π → 1). 0 (a) Write the expression for the energy of the electron orbital in terms of the speed v and the separation from the nucleus r. (b) If the maximum and minimum distance between the nucleus and the electron ( on the same orbit ) is r1 and r2 respectively, find the ratio of the speeds of these two point vv12 . (c) Find the expressions of r1 , r2 in terms of the conserved quantities involved with the orbit. (d) Using your result from the above part, show that the energy of the orbit does not depend on the eccentricity of the elliptic orbits. h (e) If the angular momentum of the orbit is given by L = k¯h where ¯h = 2π with h being the Planck’s constant, express the ratio of semi-minor axis and semi-major axis in terms of the quantum numbers involved with the orbit.

1+2+3+2+2=10 1

2. Passage of Charged Particles through Matter - 9 marks: Charged particles while passing through matter lose their energy by interacting with the electrons in the medium. The energy loss formula was first derived by Bohr around 1913 and was later given a quantum mechanical treatment by Hans Bethe. In this problem, we will retrace the steps taken for the derivation of a simplified version of the celebrated the Bethe-Bloch formula used for finding the stopping power of different materials. This physics of this process lies at the heart of most detectors used in particle physics. Consider a massive particle with mass M and charge Ze moving through a medium which has electron density n. The electrons are treated to be free and initially assumed to be at rest. Since the particle is taken to be much more massive compared to electrons , we will assume its path to be a straight line. (a) If the perpendicular distance to an electron from the path of the heavy charged particle is b, write the component of the electric force perpendicular to the path of the projectile if its velocity is v and the time passed after passing through the closest point of approach is t. (b) Using the above expressions find the change of momentum in the direction perpendicular to the path. Hint: i. you may need to use the integral Z ∞ −∞ (x2

dx +

3

a2 ) 2

=

2 a2

. ii. You can use the Gauss law to derive the same find result for this part. (c) Find the energy transferred onto a single electron as a function of the perpendicular distance b. (d) Instead of a single electron let us now consider a cloud of electrons, contained within a cylindrical shell whose symmetry axis coincides with the path of heavy charged particle. What is the number of electrons contained in this shell if its inner radius is b, has a shell thickness db and of length dx along the axis? (e) Using the above results find the energy loss per path length dE dx , if the electron is spread between a cylindrical sheel of inner radii bmin and outer radii bmax . (f) Your result should diverge ( i.e. become infinite) if bmin → 0 and/or bmax → ∞. The minimum value for bmin being zero is prevented by Heisenberg uncertainty principle. But suggest how bmax can be rendered a finite value to give rise to finite energy loss. 1+2 +1+2 +2+1=9 3. Expanding Bubbles- 6 marks: Consider a expanding gas bubble which is immersed in an incompressible liquid. Let the radius of the bubble at time t be R(t). We will also assume that the bubble remains spherical during the expansion and its center does not move with time. The liquid outside the bubble undergoes a radial motion due to the expansion of the bubble and let the radial speed of the liquid at a distance r ( which is greater than R(r)) away from the center be u(r, t) ( see the accompanying figure).

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(a) Using the incompressible nature of the liquid find the speed of the liquid u(r, t) as a ˙ function of r, R(t) and the expansion speed of the bubble dR dt ≡ R(t). (b) The kinetic energy contained in the liquid can be found by integrating the kinetic energy per unit volume of the liquid outside the bubble. Find this kinetic energy in terms of ˙ the bubble radius R(t) and the speed of bubble surface R(t). [ Hint: You will need your result from the previous part] (c) This kinetic energy is obtained by the work done by the expansion of the gas. If the pressure of the fluid at r → ∞ is p0 , find the relation between the pressure difference and bubble radius and velocity. Choose the initial bubble radius at time t = 0 to be R0 . Your result should be in the form of an integration. (d) Differentiating the above result with respect to R and using the relation ∂ R˙ 2 ¨ = 2R ∂R find the differential equation that connects the bubble acceleration and the pressure at the bubble surface. This is a simplified version of the celebrated Rayleigh-Plesset equation. 1+2+1+2=6

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Solutions to C category exam for NPhO 2015-16 BdPhO 1st Jan, 2016

Question 1 We will choose units where the Coulomb force is written as F = with a separation r.

q1 q2 r2

between charges q1 , q2

(a) The energy is simply the sum of kinetic and potential energies: 1 e2 E = mv 2 − 2 r (b) Gravitational force is central and thus exerts zero torque, which leads to conservation of angular momentum. This is nothing but the Kepler’s second law. Therefore, the angular momentum will be the same at those two points: L = mv1 r1 = mv2 r2

⇒

v1 r2 = v2 r1

(c) As L = mvr , we can eliminate v from the energy expression and rewrite it in terms of the angular momentum ( which is a conserved quantity for the orbits under Coulomb/central forces ): e2 L2 − E= 2mr2 r which can be rewritten as e2 L2 r2 + r − =0 E 2mE whose solutions are e2  e2 2 L2 1 r1,2 = − ± ( ) + ]2 2E 2E 2mE (d) The sum of these two distances is equal to the length of the major axis of the ellipse ( 2a) .

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Hence, e2 e2 ⇒ E =− E 2a which shows the independence of the energy on the minor axis length. r1 + r2 = 2a = −

(e) Let us consider when the electron passes through the point P ( see figure )

where the minor axis intersects the orbit. The angular momentum can be written as L = mv0 b. At this point, the distance between the nucleus and the electron , r0 = a ( to see this, recall that for ellipse the sum of the distance from the foci to the locus is constant. So, r1 + r2 = 2a = 2r0 ). Therefore, E=

e2 e2 mv02 e2 − =− ⇒ v02 = 2 a 2a ma

which tell us that the value of v0 , the speed at point P does not depend on b, it is fixed by a only. Once we demand the quantization of L, we see that the value of b has be discrete as well ( since L = mv0 b ). For circular orbits b = a, the latter being the maximum value for b, with the energy of the orbit, E fixed. Now, we ready see b L k~ = = a mv0 a Lmax as the angular momentum is maximum for the circular orbit ( with the energy fixed) . But Lmax = n~. This leads to our cherished result b k = a n

Question 2 We will refer to the heavy charged particle hereafter as an “ion”.

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(a) We start off with the vector form of the Coulomb’s law E=k

Ze r r3 1

Now from the figure we see that r = (b2 + v 2 t2 ) 2 . Taking the projection along the direction perpendicular to path of the “ion” , E⊥ = k

Zeb (b2

3

+ v 2 t2 ) 2

(b) The momentum transfer in the direction parallel to the path of the “ion” is zero due to the symmetry of the problem ( as the electron is initially at rest). The momentum transfer will be perpendicular to the path and will be given by Z ∞ Z ∞ p⊥ = E⊥ dt F⊥ dt = e −∞

−∞

One can now set x = vt ⇒ dt = v1 dx and use the given integral to find kZe2 b p⊥ = v

Z

∞

dx 3

(b2 + x2 ) 2

−∞

=

2kZe2 vb

Alternative: We saw above that

e p⊥ = v

Z

∞

E⊥ dx −∞

Next we imagine a Gaussian cylinder around the path of the “ion” which passes through the location of the electron ( but we do not consider the charge of the electron - as it is the test charge ) with the source charge being Ze carried by the “ion”. The infinitesimal surface area of of the Gaussian surface is given by dS = dx(2πb) so Z e e 2Ze2 k p⊥ = EdS = (4πkZe) = v(2πb 2πvb vb ( Note: For E = kq/r2 , the Gauss law takes the from ΦE = 4πkq. ) (c) The energy gained by a single electron ∆E =

2Z 2 e4 k 2 (p⊥ )2 = 2m mb2 v 2

(d) Consider a cylindrical shell of inner radius b, thickness db and axis length dx. If the charge density of n, then the number of electrons inside this cylindrical shell is dN = n(2πb) db dx which gives the number of electrons in unit thickness dN/dx = 2πnb db

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(e) The energy loss per unit length from the “ion” will therefore be given by dE dN 4πk 2 Z 2 e4 =− ∆E = − dx dx mv 2 b This expression is valid for a single shell of thickness db. Hence, the net energy loss will be found by integrating over all shells within the range [bmin , bmax ]   Z dE 4πk 2 Z 2 e4 bmax 4πk 2 Z 2 e4 bmax db =− ln =− dx mv 2 b mv 2 bmin bmin net

(f) It is obvious bmax must be finite otherwise the energy loss will be infinite. This can be justified by the fact that electric field inside matter gets screened and thus has a finite range.

Question 3 Let the radius of the bubble be R(t) which varies with time t, due to the expansion of the bubble.

(a) Let us think of a spherical surface which is concentric with the bubble with a radius r > R(t). Since the liquid outside the bubble is incompressible the volume gained by the gas bubble per unit time 4πR2 (t) dR(t) dt must match the flow of liquid through the outside surface 4πr2 u(r, t). Equating these two quantities one gets u(r, t) =

R2 ˙ R r2

(b) The kietic energy density of the liquid is h ≡ 21 ρu2 = 2rρ4 R4 R˙ 2 . Therefore, the total kinetic energy of the liquid outside is Z ∞ Z ∞ 1 T = h(4πr2 ) dr = 2πR4 R˙ 2 dr = 2πR3 (t)R˙ 2 2 r R(t) R(t) (c) The kinetic energy gained by the liquid is obtained from the work done by the pressure difference p(t) − p0 on the bubble wall. We thus get  Z R(t)  ˜ t) − p0 (4π R ˜ 2 ) dR ˜ = 2πR3 (t)R˙ 2 p(R, R0

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(d) If we differentiate both sides with respect to R(t) ≡ R, we are led to   ∂ R3 R˙ 2 1 ¨ 4πR2 (p(t) − p0 ) = 2π ⇒ 2(p(t) − p0 ) = 2 (3R2 R˙ 2 + 2R3 R) ∂R R R˙ 2 ∂ R˙ ¨ This can also be estabwhere we have employed the relation partial = 2R˙ ∂R = 2R. ∂R lished from the work-energy theorem, by differentiating both sides with respect to the displacement.

Thus, simplifying, we are led to ¨ + 3 R˙ 2 = (p(t) − p0 ) RR 2 which is a version of the Rayleigh-Plesset equation.

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