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IPhO 1983, Теория — Задача 4. Atomics - Problem IV: Compton scattering

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IPhO 1983 Theoretical Questions · 7 т.

Условие

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 $\lambda_0$ scattered at an angle $\theta = 60^{\circ}$ with respect to the direction of the incident photon, is again scattered by a second free electron at rest. In this second scattering process a photon with wavelength of $\lambda_f = 1{,}25 \times 10^{-10}\ \mathrm{m}$ emerges at an angle $\theta = 60^{\circ}$ with respect to the direction of the photon of wavelength $\lambda_0$ . Find the de Broglie wavelength for the first electron before the interaction. The following constants are known:

$h = 6{,}6 \times 10^{-34}\ \mathrm{J \cdot s}$ - Planck’s constant

$m = 9{,}1 \times 10^{-31}\ \mathrm{kg}$ - mass oh the electron

$c = 3{,}0 \times 10^{8}\ \mathrm{m/s}$ - speed of light in vacuum

The purpose of the problem is to calculate the values of the speed, momentum and wavelength of the first electron.

To characterize the photons the following notation are used:

Table 4.1

	initial photon	photon – after the first scattering	final photon
momentum	$\vec{p}_i$	$\vec{p}_0$	$\vec{p}_f$
energy	$E_i$	$E_0$	$E_f$
wavelength	$\lambda_i$	$\lambda_i$	$\lambda_f$

To characterize the electrons one uses

Table 4.2

	first electron before collision	first electron after collision	second electron before collision	Second electron after collision
momentum	$\vec{p}_{1e}$	0	0	$\vec{p}_{2e}$
energy	$E_{1e}$	$E_{0e}$	$E_{0e}$	$E_{2e}$
speed	$\vec{v}_{1e}$	0	0	$\vec{v}_{2e}$

Решение

Problem III - Solution

The purpose of the problem is to calculate the values of the speed, momentum and wavelength of the first electron.

To characterize the photons the following notation are used:

Table 4.1

	initial photon	photon – after the first scattering	final photon
momentum	$\vec p_i$	$\vec p_0$	$\vec p_f$
energy	$E_i$	$E_0$	$E_f$
wavelength	$\lambda_i$	$\lambda_i$	$\lambda_f$

To characterize the electrons one uses

Table 4.2

	first electron before collision	first electron after collision	second electron before collision	Second electron after collision
momentum	$\vec p_{1e}$	$0$	$0$	$\vec p_{2e}$
energy	$E_{1e}$	$E_{0e}$	$E_{0e}$	$E_{2e}$
speed	$\vec v_{1e}$	$0$	$0$	$\vec v_{2e}$

The image in figure 4.1 presents the situation before the first scattering of photon.

Photon/electron scattering geometry, Figure 4.1.

Photon/electron scattering geometry, Figure 4.2.

Photon/electron scattering geometry, Figure 4.3.

Photon/electron scattering geometry, Figure 4.4.

To characterize the initial photon we will use his momentum $\vec p_i$ and his energy $E_i$

\begin{cases}\vec P_i=\dfrac h{\lambda_i}=\dfrac{hf_i}c\\E_i=hf_i\end{cases}\qquad (4.1)

f_i=\frac c{\lambda_i}\qquad (4.2)

is the frequency of initial photon.

For initial, free electron in motion the momentum $\vec p_{0e}$ and the energy $E_{0e}$ are

\begin{cases}\vec P_{0e}=m\vec v_{1e}=\dfrac{m_0\vec v_{1e}}{\sqrt{1-\beta^2}}\\E_{0e}=mc^2=\dfrac{m_0c^2}{\sqrt{1-\beta^2}}\end{cases}\qquad (4.3)

where $m_0$ is the rest mass of electron and $m$ is the mass of moving electron. As usual, $\beta=v_{1e}/c$ .

De Broglie wavelength of the first electron is

\lambda_{0e}=\frac h{p_{0e}}=\frac h{m_0v_{1e}}\sqrt{1-\beta^2}.

The situation after the scattering of photon is described in the figure 4.2. To characterize the scattered photon we will use his momentum $\vec p_0$ and his energy $E_0$

\begin{cases}\vec P_0=\dfrac h{\lambda_0}=\dfrac{hf_0}c\\E_0=hf_0\end{cases}\qquad (4.4)

where

f_0=\frac c{\lambda_0}\qquad (4.5)

is the frequency of scattered photon.

The magnitude of momentum of the electron (that remains in rest) after the scattering is zero; his energy is $E_{1e}$ . The mass of electron after collision is $m_0$ - the rest mass of electron at rest. So,

E_{1e}=m_0c^2.

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_{0e}=\vec p_0\qquad (4.6)

and

E_i+E_{0e}=E_0+E_{1e}.\qquad (4.7)

The conservation of moment on $Ox$ direction is written as

\frac{hf_i}c+mv_{1e}\cos\alpha=\frac{hf_0}c\cos\theta\qquad (4.8)

and the conservation of moment on $Oy$ is

mv_{1e}\sin\alpha=\frac{hf_0}c\sin\theta.\qquad (4.9)

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

\begin{cases}(mv_{1e}\cos\alpha)^2=\dfrac{h^2}{c^2}(f_0\cos\theta-f_i)^2\\(mv_{1e}\sin\alpha)^2=\left(\dfrac{hf_0}c\sin\theta\right)^2\end{cases}\qquad (4.10)

and then added. The result is

m^2v_{1e}^2=\frac{h^2}{c^2}(f_0^2+f_1^2-2f_0f_i\cos\theta)\qquad (4.11)

\frac{m_0^2c^2}{1-(v_{1e}/c)^2}v_{1e}^2=h^2(f_0^2+f_1^2-2f_0f_i\cos\theta).\qquad (4.12)

The conservation of energy (4.7) can be written again as

mc^2+hf_1=m_0c^2+hf_0\qquad (4.13)

\frac{m_0c^2}{\sqrt{1-(v_{1e}/c)^2}}=m_0c^2+h(f_0-f_1).\qquad (4.14)

Squaring the last relation results

\frac{m_0^2c^4}{1-(v_{1e}/c)^2}=m_0^2c^4+h^2(f_0-f_1)^2+m_0hc^2(f_0-f_1).\qquad (4.15)

Subtracting (4.12) from (4.15) the result is

2m_0c^2h(f_0-f_1)+2h^2f_1f_0\cos\theta-2h^2f_1f=0\qquad (4.16)

\frac h{m_0c}(1-\cos\theta)=\frac c{f_1}-\frac c{f_0}.\qquad (4.17)

Using

\Lambda=\frac h{m_0c}\qquad (4.18)

the relation (4.17) becomes

\Lambda(1-\cos\theta)=\lambda_i-\lambda_0.\qquad (4.19)

The wavelength of scattered photon is

\lambda_0=\lambda_i-\Lambda(1-\cos\theta)\qquad (4.20)

shorter than the wavelength of initial photon and consequently the energy of scattered photon is greater that the energy of initial photon.

\begin{cases}\lambda_i<\lambda_0\\E_i>E_0\end{cases}\qquad (4.21)

Let's analyze now the second collision process that occurs in point $N$ . To study that, let's consider a new referential having $Ox$ direction on the direction of the photon scattered after the first collision.

The figure 4.3 presents the situation before the second collision and the figure 4.4 presents the situation after this scattering process. The conservation principle for moment in the scattering process gives

\begin{cases}\dfrac h{\lambda_0}=\dfrac h{\lambda_f}\cos\theta+mv_{2e}\cos\beta\\\dfrac h{\lambda_f}\sin\theta-mv_{2e}\sin\beta=0\end{cases}\qquad (4.22)

To eliminate the unknown angle $\beta$ must square and then add the equations (4.22)

That is

\begin{cases}\left(\dfrac h{\lambda_0}-\dfrac h{\lambda_f}\cos\theta\right)^2=(mv_{2e}\cos\beta)^2\\\left(\dfrac h{\lambda_f}\sin\theta\right)^2=(mv_{2e}\sin\beta)^2\end{cases}\qquad (4.23)

\left(\frac h{\lambda_f}\right)^2+\left(\frac h{\lambda_0}\right)^2-\frac{2h^2}{\lambda_0\lambda_f}\cos\theta=(mv_{2e})^2.\qquad (4.24)

The conservation principle of energy in the second scattering process gives

\frac{hc}{\lambda_0}+m_0c^2=\frac{hc}{\lambda_f}+mc^2.\qquad (4.25)

(4.24) and (4.25) gives

\frac{h^2c^2}{\lambda_f^2}+\frac{h^2c^2}{\lambda_0^2}-\frac{2h^2c^2}{\lambda_0\lambda_f}\cos\theta=m^2c^2v_{2e}^2\qquad (4.26)

and

h^2c^2\left(\frac1{\lambda_f}-\frac1{\lambda_0}\right)^2+m_0^2c^4+2hc^3m_0\left(\frac1{\lambda_f}-\frac1{\lambda_0}\right)=m^2c^4.\qquad (4.27)

Subtracting (4.26) from (1.27), one obtain

\begin{cases}\dfrac h{m_0c}(1-\cos\theta)=\lambda_f-\lambda_0\\\lambda_f-\lambda_0=\Lambda(1-\cos\theta)\end{cases}\qquad (4.28)

That is

\begin{cases}\lambda_f>\lambda_0\\E_f<E_0\end{cases}\qquad (4.29)

Because the value of $\lambda_f$ is known and $\Lambda$ can be calculate as

\begin{aligned}\lambda_f&=1.25\times10^{-10}\,\mathrm m\\\Lambda&=\frac{6.6\times10^{-34}}{9.1\times10^{-31}\cdot3\times10^8}\,\mathrm m=2.41\times10^{-12}\,\mathrm m=0.02\times10^{-10}\,\mathrm m\end{aligned}\qquad (4.30)

the value of wavelength of photon before the second scattering is

\lambda_0=1.23\times10^{-10}\,\mathrm m.\qquad (4.31)

Comparing (4.28) written as:

\lambda_f=\lambda_0+\Lambda(1-\cos\theta)\qquad (4.32)

and (4.20) written as

\lambda_i=\lambda_0+\Lambda(1-\cos\theta)\qquad (4.33)

clearly results

\lambda_i=\lambda_f.\qquad (4.34)

The energy of the double scattered photon is the same as the energy of initial photon. The direction of “final photon” is the same as the direction of “initial” photon. Concluding, the final photon is identical with the initial photon. The result is expected because of the symmetry of the processes.

Extending the symmetry analyze on electrons, the first moving electron that collides the initial photon and after that remains at rest must have the same momentum and energy as the second electron after the collision – because this second electron is at rest before the collision.

That is

\begin{cases}\vec p_{1e}=\vec p_{2e}\\E_{1e}=E_{2e}\end{cases}\qquad (4.35)

Taking into account (4.24), the moment of final electron is

p_{2e}=h\sqrt{\frac1{\lambda_f^2}+\frac1{(\lambda_f-\Lambda(1-\cos\theta))^2}-\frac{2\cos\theta}{\lambda_f(\lambda_f-\Lambda(1-\cos\theta))}}.\qquad (4.36)

The de Broglie wavelength of second electron after scattering (and of first electron before scattering) is

\lambda_{1e}=\lambda_{2e}=\frac1{\sqrt{\frac1{\lambda_f^2}+\frac1{(\lambda_f-\Lambda(1-\cos\theta))^2}-\frac{2\cos\theta}{\lambda_f(\lambda_f-\Lambda(1-\cos\theta))}}}.\qquad (4.37)

Numerical value of this wavelength is

\lambda_{1e}=\lambda_{2e}=1.24\times10^{-10}\,\mathrm m.\qquad (4.38)

Professor Delia DAVIDESCU, National Department of Evaluation and Examination–Ministry of Education and Research- Bucharest, Romania

Professor Adrian S. DAFINEI, PhD, Faculty of Physics – University of Bucharest, Romania

Marking scheme — Question 4

IV. — 7.0 points

For: the situation before the first scattering of photon

Marking-scheme photon/electron scattering geometry.

Marking scheme — The theoretical question IV — Compton scattering	Score
The momentum $\vec p_i$ and the energy $E_i$ of the initial photon: $\vec P_i=h/\lambda_i=hf_i/c$ , $E_i=hf_i$	0.3
The frequency of initial photon $f_i=c/\lambda_i$	0.1
The momentum $\vec p_{0e}$ and the energy $E_{0e}$ of initial, free electron in motion: $\vec P_{0e}=m\vec v_{1e}=m_0\vec v_{1e}/\sqrt{1-\beta^2}$ ; $E_{0e}=mc^2=m_0c^2/\sqrt{1-\beta^2}$	0.3
De Broglie wavelength of the first electron $\lambda_{0e}=h/p_{0e}=h\sqrt{1-\beta^2}/(m_0v_{1e})$	0.2

The situation after the scattering of photon

Marking scheme	Score
The momentum $\vec p_0$ and the energy $E_0$ of the scattered photon: $\vec P_0=h/\lambda_0=hf_0/c$ ; $E_0=hf_0$	0.3
The frequency of scattered photon $f_0=c/\lambda_0$	0.1
The principles of conservation of moments and energy: $\vec P_i+\vec p_{0e}=\vec p_0$ ; $E_i+E_{0e}=E_0+E_{1e}$	0.3
The conservation of moment on $Ox$ direction: $hf_i/c+mv_{1e}\cos\alpha=(hf_0/c)\cos\theta$	0.3
The conservation of moment on $Oy$ : $mv_{1e}\sin\alpha=(hf_0/c)\sin\theta$	0.3
$\dfrac{m_0^2c^2v_{1e}^2}{1-(v_{1e}/c)^2}=h^2(f_0^2+f_1^2-2f_0f_i\cos\theta)$	0.4
The conservation of energy $mc^2+hf_1=m_0c^2+hf_0$	0.3
Or $\dfrac{m_0c^2}{\sqrt{1-(v_{1e}/c)^2}}=m_0c^2+h(f_0-f_1)$	0.2
$\dfrac{m_0^2c^4}{1-(v_{1e}/c)^2}=m_0^2c^4+h^2(f_0-f_1)^2+m_0hc^2(f_0-f_1)$	0.2
$\dfrac h{m_0c}(1-\cos\theta)=c/f_1-c/f_0$	0.2
$\Lambda=h/(m_0c)$	0.2
The wavelength of scattered photon $\lambda_0=\lambda_i-\Lambda(1-\cos\theta)$	0.3
$\lambda_i<\lambda_0$ ; $E_i>E_0$	0.1

The situation before the second collision

The situation after this scattering process

Marking scheme	Score
The conservation principle for moment in the scattering process: $h/\lambda_0=(h/\lambda_f)\cos\theta+mv_{2e}\cos\beta$ ; $(h/\lambda_f)\sin\theta-mv_{2e}\sin\beta=0$	0.3
$(h/\lambda_f)^2+(h/\lambda_0)^2-\dfrac{2h^2}{\lambda_0\lambda_f}\cos\theta=(mv_{2e})^2$	0.3
$\dfrac h{m_0c}(1-\cos\theta)=\lambda_f-\lambda_0$ ; $\lambda_f-\lambda_0=\Lambda(1-\cos\theta)$	0.5
$\lambda_f>\lambda_0$ ; $E_f<E_0$	0.1
$\lambda_f=1.25\times10^{-10}\,\mathrm m$ ; $\Lambda=\dfrac{6.6\times10^{-34}}{9.1\times10^{-31}\cdot3\times10^8}\,\mathrm m=2.41\times10^{-12}\,\mathrm m=0.02\times10^{-10}\,\mathrm m$	0.2
The value of wavelength of photon before the second scattering $\lambda_0=1.23\times10^{-10}\,\mathrm m$	0.1
$\lambda_i=\lambda_f$	0.3
$\vec p_{1e}=\vec p_{2e}$ ; $E_{1e}=E_{2e}$	0.2
The moment of final electron, below	0.4

p_{2e}=h\sqrt{\frac1{\lambda_f^2}+\frac1{(\lambda_f-\Lambda(1-\cos\theta))^2}-\frac{2\cos\theta}{\lambda_f(\lambda_f-\Lambda(1-\cos\theta))}}.

The de Broglie wavelength of second electron after scattering (and of first electron before scattering)

\lambda_{1e}=\lambda_{2e}=\frac1{\sqrt{\frac1{\lambda_f^2}+\frac1{(\lambda_f-\Lambda(1-\cos\theta))^2}-\frac{2\cos\theta}{\lambda_f(\lambda_f-\Lambda(1-\cos\theta))}}}.

Marking scheme	Score
De Broglie wavelength expression above	0.3
Final result: $\lambda_{1e}=\lambda_{2e}=1.24\times10^{-10}\,\mathrm m$	0.2
Total score theoretical question IV	7.0 points