Съдържание

IPhO 2015, Теория — Задача 3.2. The Moderator

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Всички материали от IPhO 2015

The Design of a Nuclear Reactor

Footnote 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.

Условие

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 $\overrightarrow{v_b}$ and $\overrightarrow{v_a}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_m}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds.

B1 The collision in LF is shown schematically, where $\theta_L$ 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 $\overrightarrow{v_b}$ , $\overrightarrow{v_a}$ and $\overrightarrow{v_m}$ . Indicate the scattering angle $\theta$ . [1 т.]

Fig-IV, collision in the laboratory frame: a small circle 1 with velocity v_b moves right towards a larger circle 3 at rest; after the collision the small circle 2 moves up-right with velocity v_a at angle θ_L to the dashed initial direction, and the large circle 4 moves down-right. Legend box: Collision in the Laboratory Frame; 1-Neutron before collision; 2-Neutron after collision; 3-Moderator Atom before collision; 4-Moderator Atom after collision. — Fig-IV

B2 Obtain $v$ and $V$ , the speeds of the neutron and moderator atom in the CM frame after collision, in terms of $A$ and $v_b$ . [1 т.]

B3 Derive an expression for $G(\alpha, \theta) = E_a/E_b$ , where $E_b$ and $E_a$ are the kinetic energies of the neutron, in the LF, before and after the collision respectively and $\alpha \equiv [(A-1)/(A+1)]^2$ . [1 т.]

B4 Assume that the above expression holds for D $_2$ O molecule. Calculate the maximum possible fractional energy loss $f_l \equiv \frac{E_b - E_a}{E_b}$ of the neutron for the D $_2$ O (20 u) moderator. [0,5 т.]

Отговори

Решение

Two sketches. Laboratory Frame: neutron 1 with velocity v_b hits the moderator atom 3 at rest; the neutron 2 leaves with velocity v_a at angle θ_L above the dashed initial direction and the moderator atom 4 moves down-right. Center of Mass Frame: the neutron with velocity v_b − v_m moves right and the moderator atom with velocity −v_m moves left; after the collision the neutron moves up with velocity v_a − v_m at angle θ to the initial line and the moderator atom moves in the opposite direction, down. — Laboratory Frame / Center of Mass Frame

Solution:

Solution: Detailed solution: Before the collision in the CM frame $(v_b - v_m)$ and $v_m$ will be magnitude of the velocities of the neutron and moderator atom respectively. From momentum conservation in the CM frame, $v_b - v_m = Av_m$ gives $v_m = \frac{v_b}{A+1}$ .

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 \qquad \text{and} \qquad \frac{1}{2}(v_b - v_m)^2 + \frac{1}{2}Av_m^2 = \frac{1}{2}v^2 + \frac{1}{2}AV^2.(\rightarrow [0.2 + 0.2])

Solving gives $v = \frac{Av_b}{A+1}$ and $V = \frac{v_b}{A+1}$ . (OR) From definition of center of mass frame $v_m = \frac{v_b}{A+1}$ . Before the collision in the CM frame $v_b - v_m = \frac{Av_b}{A+1}$ and $v_m$ will be magnitude 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 the speeds remain same $v = \frac{Av_b}{A+1}$ and $V = \frac{v_b}{A+1}$ $(\rightarrow [0.2 + 0.1])$ .

Note: Alternative solutions are worked out in the end and will get appropriate weightage.

Solution:

G(\alpha,\theta) = \frac{E_a}{E_b} = \frac{A^2 + 2A\cos\theta + 1}{(A+1)^2} = \frac{1}{2}\left[(1+\alpha) + (1-\alpha)\cos\theta\right].

Detailed solution: Since $\overrightarrow{v_a} = \overrightarrow{v} + \overrightarrow{v_m}$ , $v_a^2 = v^2 + v_m^2 + 2vv_m\cos\theta$ $(\rightarrow [0.3])$ . Substituting the values of $v$ and $v_m$ , $v_a^2 = \frac{A^2v_b^2}{(A+1)^2} + \frac{v_b^2}{(A+1)^2} + \frac{2Av_b^2}{(A+1)^2}\cos\theta$ $(\rightarrow [0.2])$ , so

\frac{v_a^2}{v_b^2} = \frac{E_a}{E_b} = \frac{A^2 + 2A\cos\theta + 1}{(A+1)^2}.

G(\alpha,\theta) = \frac{A^2+1}{(A+1)^2} + \frac{2A}{(A+1)^2}\cos\theta = \frac{1}{2}\left[(1+\alpha) + (1-\alpha)\cos\theta\right].

Alternate form

= 1 - \frac{(1-\alpha)(1-\cos\theta)}{2}.

Note: Alternative solutions are worked out in the end and will get appropriate weightage.

Solution: $f_l = 0.181$

Detailed solution: The maximum energy loss will be when the collision is head on ie., $E_a$ will be minimum for the scattering angle $\theta = \pi$ .

So $E_a = E_{min} = \alpha E_b$ .

For D $_2$ O, $\alpha = 0.819$ and maximum fractional loss $\left(\frac{E_b - E_{min}}{E_b}\right) = 1 - \alpha = 0.181$ . [Acceptable Range (0.170 to 0.190)]

Alternative Solutions to sub-parts B2 and B3: Let $\sigma$ 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

v_b = v_a\cos\theta_L + AU\cos\sigma, \qquad (1)

0 = v_a\sin\theta_L - AU\sin\sigma, \qquad (2)

\frac{1}{2}v_b^2 = \frac{1}{2}AU^2 + \frac{1}{2}v_a^2. \qquad (3)

Squaring and adding eq(1) and (2) to eliminate $\sigma$ and from eq(3) we get

A^2U^2 = v_a^2 + v_b^2 - 2v_av_b\cos\theta_L,

A^2U^2 = Av_b^2 - Av_a^2, \qquad (4)

which gives

2v_av_b\cos\theta_L = (A+1)v_a^2 - (A-1)v_b^2. \qquad (5)

(ii) Let $v$ be the speed of the neutron after collision in the COMF. From definition of center of mass frame $v_m = \dfrac{v_b}{A+1}$ .

$v_a\sin\theta_L$ and $v_a\cos\theta_L$ are the perpendicular and parallel components of $v_a$ , in the LF, resolved along the initial direction of the neutron before collision. Transforming these to the COMF gives $v_a\sin\theta_L$ and $v_a\cos\theta_L - v_m$ as the perpendicular and parallel components of $v$ . Substituting for $v_m$ and for $2v_av_b\cos\theta_L$ from eq(5) in $v = \sqrt{v_a^2\sin^2\theta_L + v_a^2\cos^2\theta_L + v_m^2 - 2v_av_m\cos\theta_L}$ and simplifying gives $v = \dfrac{Av_b}{A+1}$ . Squaring the components of $v$ to eliminate $\theta_L$ gives $v_a^2 = v^2 + v_m^2 + 2vv_m\cos\theta$ . Substituting for $v$ and $v_m$ and simplifying gives,

\frac{v_a^2}{v_b^2} = \frac{E_a}{E_b} = \frac{A^2 + 2A\cos\theta + 1}{(A+1)^2}.

G(\alpha,\theta) = \frac{E_a}{E_b} = \frac{A^2+1}{(A+1)^2} + \frac{2A}{(A+1)^2}\cos\theta = \frac{1}{2}\left[(1+\alpha) + (1-\alpha)\cos\theta\right].

(OR)

(iii) From definition of center of mass frame $v_m = \dfrac{v_b}{A+1}$ . After the collision, let $v$ and $V$ be magnitude of the velocities of neutron and moderator atom respectively in the COMF. From conservation laws in the COMF,

v = AV \qquad \text{and} \qquad \frac{1}{2}(v_b - v_m)^2 + \frac{1}{2}Av_m^2 = \frac{1}{2}v^2 + \frac{1}{2}AV^2.

Solving gives $v = \frac{Av_b}{A+1}$ and $V = \frac{v_b}{A+1}$ . We also have $v\cos\theta = v_a\cos\theta_L - v_m$ , substituting for $v_m$ and for $v_a\cos\theta_L$ from eq(5) and simplifying gives

\frac{v_a^2}{v_b^2} = \frac{E_a}{E_b} = \frac{A^2 + 2A\cos\theta + 1}{(A+1)^2}.

G(\alpha,\theta) = \frac{E_a}{E_b} = \frac{A^2+1}{(A+1)^2} + \frac{2A}{(A+1)^2}\cos\theta = \frac{1}{2}\left[(1+\alpha) + (1-\alpha)\cos\theta\right].

(OR)

(iv) From definition of center of mass frame $v_m = \dfrac{v_b}{A+1}$ . 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 in the CM frame,

v = AV \qquad \text{and} \qquad \frac{1}{2}(v_b - v_m)^2 + \frac{1}{2}Av_m^2 = \frac{1}{2}v^2 + \frac{1}{2}AV^2.

Solving gives $v = \frac{Av_b}{A+1}$ and $V = \frac{v_b}{A+1}$ . $U\sin\sigma$ and $U\cos\sigma$ 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 $U\sin\sigma$ and $-U\cos\sigma + v_m$ as the perpendicular and parallel components of $V$ . So we get $U^2 = V^2\sin^2\theta + V^2\cos^2\theta + v_m^2 - 2Vv_m\cos\theta$ . Since $V = v_m$ we get $U^2 = 2v_m^2(1 - \cos\theta)$ . Substituting for $U$ from eq(4) and simplifying gives

\frac{v_a^2}{v_b^2} = \frac{E_a}{E_b} = \frac{A^2 + 2A\cos\theta + 1}{(A+1)^2}.

G(\alpha,\theta) = \frac{E_a}{E_b} = \frac{A^2+1}{(A+1)^2} + \frac{2A}{(A+1)^2}\cos\theta = \frac{1}{2}\left[(1+\alpha) + (1-\alpha)\cos\theta\right].

Note: We have $v_a = \dfrac{\sqrt{A^2 + 2A\cos\theta + 1}}{A+1}v_b$ . Substituting for $v_a,\ v,\ v_m$ in $v\cos\theta = v_a\cos\theta_L - v_m$ gives the relation between $\theta_L$ and $\theta$ ,

\cos\theta_L = \frac{A\cos\theta + 1}{\sqrt{A^2 + 2A\cos\theta + 1}}.

Treating the above equation as quadratic in $\cos\theta$ gives,

\cos\theta = \frac{-\sin^2\theta_L \pm \cos\theta_L\sqrt{A^2 - \sin^2\theta_L}}{A}.

For $\theta_L = 0^{\circ}$ the root with the negative sign gives $\theta = 180^{\circ}$ which is not correct so,

\cos\theta = \frac{\cos\theta_L\sqrt{A^2 - \sin^2\theta_L} - \sin^2\theta_L}{A}.

Substituting the above expression for $\cos\theta$ in the expression for $\dfrac{v_a^2}{v_b^2}$ gives an expression in terms of $\cos\theta_L$

\frac{v_a^2}{v_b^2} = \frac{E_a}{E_b} = \frac{A^2 + 2\cos\theta_L\sqrt{A^2 - \sin^2\theta_L} + \cos2\theta_L}{(A+1)^2}.

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf