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RMPh 2013, Теория — Задача 2. Compton scattering

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Romanian Master of Physics 2013 — Theoretical Problem No. 2 (10 points) · 10 т.

Условие

An electron storage ring contains circulating electrons with high energy produced by an accelerator and rotating in an appropriate magnetic field. X - ray photons are directed so as to collide with electrons stored in storage ring. Phenomenon that occurs is known as Inverse Compton Scattering.

A photon of wavelength $\lambda_i$ is scattered by a moving, free electron. As a result the electron stops and the resulting photon of wavelength $\lambda_0$ is scattered at an angle $\theta = 60^{\circ}$ with respect to the direction of the incident photon; this photon is again scattered by a second free electron at rest. In this second scattering process a photon with wavelength of $\lambda_f = 1{,}25\times10^{-10}\,\mathrm{m}$ emerges at an angle $\theta = 60^{\circ}$ with respect to the direction of the photon of wavelength $\lambda_0$ .

To characterize the photons and the electrons during the processes use the following notation:

	initial photon	photon – after the first scattering	final photon		first electron before collision	first electron after collision	second electron before collision	Second electron after collision
moment	$\vec{p}_i$	$\vec{p}_0$	$\vec{p}_f$	moment	$\vec{p}_{1e}$	0	0	$\vec{p}_{2e}$
energy	$E_i$	$E_0$	$E_f$	energy	$E_{1e}$	$E_{0e}$	$E_{0e}$	$E_{2e}$
wavelength	$\lambda_i$	$\lambda_0$	$\lambda_f$	speed	$\vec{v}_{1e}$	0	0	$\vec{v}_{2e}$

The following constants are known:

$h = 6{,}6\times10^{-34}\,\mathrm{J\cdot s}$ - Planck’s constant

$m_0 = 9{,}1\times10^{-31}\,\mathrm{kg}$ - rest mass of the electron

$c = 3{,}0\times10^{8}\,\mathrm{m/s}$ - speed of light in vacuum

Task No. 1 - First collision

1.a. Draw simple sketches marking the moments of the electron and photon before and after the first collision. Clearly specify the coordinate system used. [0,5 т.]

1.b. Express the energy and moment of the electron implied in the first collision as a function of the initial speed of the electron $\vec{v}_{1e}$ and its rest mass $m_0$ . [0,8 т.]

1.c. Express the energy and wavelength of the photon after the first collision as a function of the wavelength of the initial photon $\lambda_i$ , the scattering angle $\theta$ and $\Lambda = h/(m_0\cdot c)$ . [1,2 т.]

Task No. 2 - Second collision

2.a. Draw simple sketches marking moment of the electron and photon before and after the second collision. Clearly specify the coordinate system used. [0,5 т.]

2.b. Express the energy and wavelength of the photon after the second collision as a function of the wavelength of the photon before the collision $\lambda_0$ , the scattering angle $\theta$ and $\Lambda = h/(m_0\cdot c)$ . [1,2 т.]

2.c. Express the kinetic energy $(T_2 = E_{2e} - E_{0e})$ and momentum $p_{2e}$ of the electron after the second collision as a function of the photon wavelengths after the collision $\lambda_f$ , $m_0$ , $c$ and $h$ . [0,8 т.]

Task No. 3 - Quantitative description of processes

Using the values of the given physical constants and numerical values of $\lambda_f$ and $\theta$ determine the expressions and numerical values for:

3.a. the De Broglie wavelength of the initial electron; [2 т.]

3.b. the energy and frequency of the initial photon; [1 т.]

3.c. the speed of the second electron after collision; [1 т.]

3.d. the variation of the photon wavelength after each collision process. [1 т.]

© The Problem is proposed by:

Delia DAVIDESCU, PhD

Adrian DAFINEI, PhD

Отговори

Решение

Compton scattering - Solution

Task 1 - First collision

1.a. Figure 1 presents the situation before the first scattering of the photon.

Sketch before the first collision: coordinate axes x, y with origin M in the upper left; an initial photon (wavy line marked i) travels along a horizontal line with momentum p_i towards point M; the first electron (black dot) with momentum p_1e moves up-left along a slanted line through M, making angle α with the horizontal. — Figure 1 Photon and electron before the first collision. The collision occurs in the point $M$

Sketch after the first collision: the electron at rest (black dot) at point M on a slanted line; the scattered photon (wavy line) leaves M with momentum p_o at angle θ above the horizontal; coordinate axes x, y with origin M below. — Figure 2. Photon and electron after the first collision. The collision took place in point $M$

1.b. The angle between the directions of the electron and photon moments before collision is denoted by $\alpha$

To characterize the initial photon we will use its moment $\vec{p}_i$ and its energy $E_i$

\begin{cases} |\vec{p}_i| = \dfrac{h}{\lambda_i} = \dfrac{h\cdot f_i}{c} \\ E_i = h\cdot f_i \end{cases} \qquad (1)

f_i = \frac{c}{\lambda_i} \qquad (2)

is the frequency of the initial photon.

For the initial free electron in motion scenario, the moment $\vec{p}_{1e}$ and the energy $E_{oe}$ are

\begin{cases} \vec{p}_{1e} = m\cdot\vec{v}_{1e} = \dfrac{m_0\cdot\vec{v}_{1e}}{\sqrt{1-\beta^2}} \\ E_{1e} = m\cdot c^2 = \dfrac{m_0\cdot c^2}{\sqrt{1-\beta^2}} \end{cases} \qquad (3)

where $m_0$ is the rest mass of the electron and $m$ is the mass of the moving electron. As usual $\beta = \dfrac{v_{1e}}{c}$ . De Broglie wavelength of the first electron is

\lambda_{1e} = \frac{h}{p_{1e}} = \frac{h\cdot}{m_0\cdot v_{1e}}\sqrt{1-\beta^2} \qquad (4)

1.c. The situation after the scattering of photon is described in the figure 2.

To characterize the scattered photon we will use its moment $\vec{p}_0$ and its energy $E_0$

\begin{cases} |\vec{p}_o| = \dfrac{h}{\lambda_o} = \dfrac{h\cdot f_o}{c} \\ E_o = h\cdot f_o \end{cases} \qquad (5)

where

f_o = \frac{c}{\lambda_0} \qquad (6)

is the frequency of the scattered photon.

The magnitude of the moment of the electron (that remains at rest) after the scattering is zero; its energy is $E_{0e}$ . The mass of the electron after collision is $m_0$ - the mass of electron at rest state. So,

E_{0e} = m_0\cdot c^2 \qquad (7)

To determine the moment of the first moving electron, one can write the principles of conservation of moments and energy. That is

\vec{p}_i + \vec{p}_{1e} = \vec{p}_0 \qquad (8)

and

E_i + E_{1e} = E_0 + E_{0e} \qquad (9)

Using the referential in figures 1 and 2, the conservation of the moment in collision along the $Ox$ direction is written as

\frac{h\cdot f_i}{c} + m\cdot v_{1e}\cdot\cos\alpha = \frac{h\cdot f_0}{c}\cos\theta \qquad (10)

and the conservation of moment along $Oy$ direction is

m\cdot v_{1e}\cdot\sin\alpha = \frac{h\cdot f_0}{c}\sin\theta \qquad (11)

To eliminate $\alpha$ , the last two equations must be written again as

\begin{cases} (m\cdot v_{1e}\cdot\cos\alpha)^2 = \dfrac{h^2\cdot}{c^2}(f_0\cdot\cos\theta - f_i)^2 \\ (m\cdot v_{1e}\cdot\sin\alpha)^2 = \left(\dfrac{h\cdot f_0}{c}\sin\theta\right)^2 \end{cases} \qquad (12)

and then added.

The result is

m^2\cdot v_{1e}^2 = \frac{h^2\cdot}{c^2}\left(f_0^2 + f_i^2 - 2f_0\cdot f_i\cdot\cos\theta\right) \qquad (13)

\frac{m_0^2\cdot c^2}{1-\left(\dfrac{v_{1e}}{c}\right)^2}\cdot v_{1e}^2 = h^2\cdot\left(f_0^2 + f_i^2 - 2f_0\cdot f_i\cdot\cos\theta\right) \qquad (14)

The conservation of energy (9) can be written again as

m\cdot c^2 + h\cdot f_i = m_0\cdot c^2 + h\cdot f_0 \qquad (15)

\frac{m_0\cdot c^2}{\sqrt{1-\left(\dfrac{v_{1e}}{c}\right)^2}} = m_0\cdot c^2 + h\cdot(f_0 - f_i) \qquad (16)

Squaring the last relation results in the following

\frac{m_0^2\cdot c^4}{1-\left(\dfrac{v_{1e}}{c}\right)^2} = m_0^2\cdot c^4 + h^2\cdot(f_0 - f_i)^2 + 2m_0\cdot h\cdot c^2\cdot(f_0 - f_i) \qquad (17)

Subtracting (14) from (17) the result is

2m_0\cdot c^2\cdot h\cdot(f_0 - f_i) + 2h^2\cdot f_i\cdot f_0\cdot\cos\theta - 2h^2\cdot f_i\cdot f_0 = 0 \qquad (18)

\frac{h}{m_0\cdot c}(1-\cos\theta) = \frac{c}{f_i} - \frac{c}{f_0} \qquad (19)

Using

\Lambda = \frac{h}{m_0\cdot c} \qquad (20)

the relation (19) becomes

\Lambda\cdot(1-\cos\theta) = \lambda_i - \lambda_0 \qquad (21)

The wavelength of scattered photon is

\lambda_0 = \lambda_i - \Lambda\cdot(1-\cos\theta) \qquad (22)

Equation (22) show that the scattered photon wavelength is shorter than the wavelength of the initial photon. Consequently the energy of the scattered photon is greater that the energy of the initial photon.

\begin{cases} \lambda_0 \lt \lambda_i \\ E_0 \gt E_i \end{cases} \qquad (23)

Task 2 - Second collision

2.a. Let’s analyze now the second collision process that occurs at point $N$ . To study that, let’s consider a new referential having the $Ox$ axis along the direction of the photon scattered after the first collision.

Sketch before the second collision: rotated coordinate axes x (up-right) and y (up-left) with origin N; the scattered photon (wavy line) with momentum p_0 travels up-right along the x direction towards an electron at rest (black dot) at point N. — Figure 3 Photon and electron before the second collision. The collision occur at the point $N$

Figure 3 presents the situation before the second collision and Figure 4 presents the situation after this scattering process.

Sketch after the second collision at point N: the photon scattered again (wavy line marked f) leaves horizontally to the right with momentum p_f at angle θ from the incoming direction; the recoiling electron (black dot) moves up-left with momentum p_2e at angle β; rotated coordinate axes x, y with origin N below. — Figure 4 Photon and electron after the second collision. The collision took place at the point $N$ . The scattering angle for electron is denoted by $\beta$ .

2.b. The conservation law for moments in the scattering process gives

\begin{cases} \dfrac{h}{\lambda_0} = \dfrac{h}{\lambda_f}\cos\theta + m\cdot v_{2e}\cdot\cos\beta \\ \dfrac{h}{\lambda_f}\sin\theta - m\cdot v_{2e}\cdot\sin\beta = 0 \end{cases} \qquad (24)

To eliminate the unknown angle $\beta$ , one must square and then add equations ( 24)

That is

\begin{cases} \left(\dfrac{h}{\lambda_0} - \dfrac{h}{\lambda_f}\cos\theta\right)^2 = (m\cdot v_{2e}\cdot\cos\beta)^2 \\ \left(\dfrac{h}{\lambda_f}\sin\theta\right)^2 = (m\cdot v_{2e}\cdot\sin\beta)^2 \end{cases} \qquad (25)

and

\left(\frac{h}{\lambda_f}\right)^2 + \left(\frac{h}{\lambda_0}\right)^2 - \frac{2\cdot h^2}{\lambda_0\cdot\lambda_f}\cos\theta = (m\cdot v_{2e})^2 \qquad (26)

\frac{h^2\cdot c^2}{\lambda_f^2} + \frac{h^2\cdot c^2}{\lambda_0^2} - \frac{2\cdot h^2\cdot c^2}{\lambda_0\cdot\lambda_f}\cos\theta = m^2\cdot c^2\cdot v_{2e}^2 \qquad (27)

The conservation law for energies in the second scattering process gives

\frac{h\cdot c}{\lambda_0} + m_0\cdot c^2 = \frac{h\cdot c}{\lambda_f} + m\cdot c^2 \qquad (28)

h^2\cdot c^2\cdot\left(\frac{1}{\lambda_f} - \frac{1}{\lambda_0}\right)^2 + m_0^2\cdot c^4 + 2h\cdot c^3\cdot m_0\cdot\left(\frac{1}{\lambda_f} - \frac{1}{\lambda_0}\right) = \frac{m_0^2\cdot c^4}{1-(v_{2e}/c)^2} \qquad (29)

Subtracting (27) from (29), one obtains

+\frac{2\cdot h^2\cdot c^2}{\lambda_0\cdot\lambda_f}\cos\theta - \frac{2\cdot h^2\cdot c^2}{\lambda_0\cdot\lambda_f} + 2h\cdot c^3\cdot m_0\cdot\left(\frac{1}{\lambda_f} - \frac{1}{\lambda_0}\right) = 0 \qquad (30)

that is

\begin{cases} \dfrac{h}{m_0\cdot c}\cdot(1-\cos\theta) = \lambda_f - \lambda_0 \\ \lambda_f - \lambda_0 = \Lambda\cdot(1-\cos\theta) \end{cases} \qquad (31)

Concluding

\begin{cases} \lambda_f \gt \lambda_0 \\ E_f \lt E_0 \end{cases} \qquad (32)

Comparing (31) written as

\lambda_f = \lambda_0 + \Lambda\cdot(1-\cos\theta) \qquad (33)

and (22) written as

\lambda_i = \lambda_0 + \Lambda\cdot(1-\cos\theta) \qquad (34)

results in

\lambda_i = \lambda_f \qquad (35)

2.c. Taking into account (26), the moment of the electron after the second collision is

p_{2e} = h\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{\lambda_0^2} - \frac{2\cdot\cos\theta}{\lambda_f\cdot\lambda_0}} \qquad (36)

Or, considering (33), the moment can be rewritten as

p_{2e} = h\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{(\lambda_f - \Lambda(1-\cos\theta))^2} - \frac{2\cdot\cos\theta}{\lambda_f\cdot(\lambda_f - \Lambda(1-\cos\theta))}} \qquad (37)

The expression of the de Broglie wavelength of the second electron after scattering is

\lambda_{2e} = 1\Big/\left(\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{(\lambda_f - \Lambda(1-\cos\theta))^2} - \frac{2\cdot\cos\theta}{\lambda_f\cdot(\lambda_f - \Lambda(1-\cos\theta))}}\right) \qquad (38)

From (28) results

\frac{h\cdot c}{\lambda_0} = \frac{h\cdot c}{\lambda_f} + T_2 \qquad (39)

that is

T_2 = h\cdot c\cdot\left(\frac{1}{\lambda_0} - \frac{1}{\lambda_f}\right) \qquad (40)

Task 3 - Quantitative description of processes

3.a. The relation (13) can be written as

p_{1e} = h\sqrt{\frac{1}{\lambda_i^2} + \frac{1}{\lambda_0^2} - \frac{2\cdot\cos\theta}{\lambda_i\cdot\lambda_0}} \qquad (41)

or, considering (34) and (35)

p_{1e} = h\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{(\lambda_f - \Lambda(1-\cos\theta))^2} - \frac{2\cdot\cos\theta}{\lambda_f\cdot(\lambda_f - \Lambda(1-\cos\theta))}} \qquad (42)

so that the expression of the de Broglie wavelength of the first electron before scattering is

\lambda_{1e} = 1\Big/\left(\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{(\lambda_f - \Lambda(1-\cos\theta))^2} - \frac{2\cdot\cos\theta}{\lambda_f\cdot(\lambda_f - \Lambda(1-\cos\theta))}}\right) \qquad (43)

In the condition of the problem (43) becomes

\lambda_{1e} = 1\Bigg/\left(\sqrt{\frac{1}{\lambda_f^2} + \frac{1}{\left(\lambda_f - \dfrac{\Lambda}{2}\right)^2} - \frac{1}{\lambda_f\cdot\left(\lambda_f - \dfrac{\Lambda}{2}\right)}}\right) \qquad (44)

Because the value of $\lambda_f$ and $\Lambda$ are known as

\begin{cases} \lambda_f = \lambda_i = 1{,}25\times10^{-10}\,\mathrm{m} \\ \Lambda = \dfrac{6{,}6\times10^{-34}}{9{,}1\times10^{-31}\cdot3{,}0\times10^{8}}\,\mathrm{m} = 2{,}41\times10^{-12}\,\mathrm{m} = 0{,}02\times10^{-10}\,\mathrm{m} \end{cases} \qquad (45)

From (33) results that

\lambda_0 = \lambda_f - \Lambda\cdot(1-\cos\theta) = \lambda_f - \frac{\Lambda}{2} \qquad (46)

\lambda_0 = 1{,}24\times10^{-10}\,\mathrm{m} \qquad (47)

The numerical value of the de Broglie wavelength of the first electron before scattering is

\lambda_{1e} = 10^{-10}\Big/\left(\sqrt{\frac{1}{1{,}25^2} + \frac{1}{1{,}24^2} - \frac{1}{1{,}25\times1{,}24}}\right)\mathrm{m} \cong 1{,}25\times10^{-10}\,\mathrm{m} \qquad (48)

Evidently the de Broglie wavelength $\lambda_{2e}$ for the second electron after the scattering is the same

\lambda_{1e} = \lambda_{2e} \qquad (49)

3.b. Because $\lambda_f = \lambda_i$ the energy of the first photon before the first scattering has the following expression

E_i = \frac{h\cdot c}{\lambda_i} = \frac{h\cdot c}{\lambda_f} \qquad (50)

Its numerical value is

E_i = \frac{h\cdot c}{\lambda_i} = \frac{6{,}6\times10^{-34}\cdot3{,}0\times10^{8}}{1{,}25\times10^{-10}} \cong 1{,}58\times10^{-15}\,\mathrm{J} = 9{,}9\,\mathrm{KeV} \qquad (51)

The frequency $f_i$ of the first photon before the first scattering has the following expression

f_i = \frac{c}{\lambda_i} = \frac{c}{\lambda_f} \qquad (52)

Its numerical value is

f_i = \frac{3\times10^{8}}{1{,}25\times10^{-10}} \cong 2{,}4\times10^{18}\,\mathrm{Hz} \qquad (53)

3.c. From (40) written as

T_2 = T_1 = m\cdot c^2 - m_0\cdot c^2 = h\cdot c\cdot\left(\frac{1}{\lambda_0} - \frac{1}{\lambda_f}\right) \qquad (54)

results

m = m_0 + \frac{h}{c}\cdot\left(\frac{1}{\lambda_f - (\Lambda/2)} - \frac{1}{\lambda_f}\right) \cong m_0 + \frac{h}{c}\cdot\frac{\Lambda}{2\lambda_f^2} = m_0\cdot\left(1 + \frac{1}{2}\left(\frac{h}{m_0\cdot c\cdot\lambda_f}\right)^2\right) \qquad (55)

Its numerical value is

m \cong m_0 \qquad (56)

Consequently, the speed of the second electron after the collision $v_{2e}$ has the following expression

v_{2e} = \frac{h}{\lambda_f\cdot m} \approx \frac{h}{\lambda_f\cdot m_0} \qquad (57)

Its numerical value is

v_{2e} \approx 5{,}8\times10^{6}\,\mathrm{m/s} = 0{,}019\cdot c \qquad (58)

3.d. As results from (46)

\lambda_i - \lambda_0 = \lambda_f - \lambda_0 = \frac{\Lambda}{2} \qquad (59)

The numerical value of the variation of wavelength is

\frac{\Lambda}{2} \cong 0{,}01\times10^{-10}\,\mathrm{m} \qquad (60)

Delia DAVIDESCU, PhD

Adrian DAFINEI, PhD

ANSWER SHEET

Theoretical Problem No. 2 (10 points)

Compton scattering

Task	Answer-sheet prompt	Points
1.a.	Draw simple sketches marking the moments of the electron and photon before and after the first collision. Clearly specify the coordinate system used.	0,50p
1.b.	Express the energy and moment of the electron implied in the first collision as a function of the initial speed of the electron $\vec{v}_{1e}$ and its rest mass $m_0$ .	0,80p
1.c.	Express the energy and wavelength of the photon after the first collision as a function of the wavelength of the initial photon $\lambda_i$ , the scattering angle $\theta$ and $\Lambda = h/(m_0\cdot c)$ .	1,20p
2.a.	Draw simple sketches marking moment of the electron and photon before and after the second collision. Clearly specify the coordinate system used.	0,50p
2.b.	Express the energy and wavelength of the photon after the second collision as a function of the wavelength of the photon before the collision $\lambda_0$ , the scattering angle $\theta$ and $\Lambda = h/(m_0\cdot c)$ .	1,20p
2.c.	Express the kinetic energy $(T_2 = E_{2e} - E_{0e})$ and momentum $p_{2e}$ of the electron after the second collision as a function of the photon wavelengths after the collision $\lambda_f$ , $m_0$ , $c$ and $h$ .	0,80p
3.a.	Determine the expression for the de Broglie wavelength of the initial electron	1,00p
	Determine the numerical value for the de Broglie wavelength of the initial electron	1,00p
3.b.	Determine the expressions for the energy and for the frequency of the initial photon	0,50p
	Determine the numerical values for the energy and for the frequency of the initial photon	0,50p
3.c.	Determine the expression for the speed of the second electron after collision	0,50p
	Determine the numerical value for the speed of the second electron after collision	0,50p
3.d.	Determine the expression for the variation of the photon wavelength after each collision process	0,50p
	Determine the numerical value for the variation of the photon wavelength after each collision process	0,50p

Оригинал в Архива: Theoretical_Problem_2_Question.pdf · официални решения: Theoretical_Problem_2_Solution.pdf