Exercise 1: In a Compton scattering experiment, an x-ray photon scatters through
ID: 1495738 • Letter: E
Question
Exercise 1: In a Compton scattering experiment, an x-ray photon scatters through an angle of 13.0° from a free electron that is initially at rest. The electron recoils with a speed of 1,720 km/s.
(a) Calculate the wavelength of the incident photon.___________ nm
(b) Calculate the angle through which the electron scatters._____________ degrees
Exercise 2: An electron has a kinetic energy of 3.85 eV. Find its wavelength. __________ nm
Exercise 3: (a) If the wavelength of an electron is 5.20 107 m, how fast is it moving? __________ km/s (b) If the electron has a speed equal to 6.60 106 m/s, what is its wavelength? ____________ m
Explanation / Answer
Given Data
Scattering angle of the photon, = 130
Speed of the electron, v = 1720 km/s
= (1720 km/s)(103m / 1 km)
= 1720 *103m/s
Let the wavelength of incident photon be and scattered photon be '
Solution:
(a) According to the conservation of energy
h c / = h c / ' + 1/2 mev2
(6.63 *10-34J.s)(3.0 *108m/s) / = (6.63 *10-34J.s)(3.0 *108m/s) / '+
1/2 (9.11 *10-31kg) (1720 *103m/s)2
[1/ - 1/ ' ] = 5.8253 *106
1/ ' = 1/ - 5.8253 *106 ...... (1)
-------------------------------------------------------------------------------------------------------
The compton shift is,
' - = h / mec ( 1 - cos)
' = [(6.63 *10-34J.s) / (9.11 *10-31kg)(3.0 *108m/s) ](1 - cos13) +
= 6.21758 *10-14+ ...... (2)
Substitute the value of ' in equation (1),
1/ (6.21758 *10-14+ ) = 1/ - 5.8253 *106
On solving this equation, we get 5.8253*1062+ 3.6219 *10-7 - 6.21758 *10-14= 0
= 1.033 *10-10m or -1.0328*10^-10 m---------------------------------------------------------------------------------------------------------- (b) From conservation of momentum principle we have, p ' sin = pe sin h/' sin (11.8) = me v sin ...... (3)
From equation (2),
' = 6.21758 *10-14+
= 6.21758 *10-14+ 1.0328*10-10m
= 1.033*10-9m
Now, taking equation (3),
(6.63 *10-34J.s) /(1.033 *10-9m) sin 13 = (9.11 *10-31kg)(1720 *103m/s) sin
= 0.630
Excercise 2)
K.E. = hf = h*c/lambda
lambda = h*c / K.E. = 6.626 X 10^-34 * 3 X 10^8 / 3.85 X 1.6 X 10^-19 = 3.2269 x 10^-7 = 322.69 nm = 322.7 nm
Excercise 3)
a)
1/2*m*v^2 = h*c/lambda
v = sqrt(2*h*c/ m*lambda) = sqrt(2*6.626 x 10^-34 * 3x 10^8 / 9.1 x 10^-31 *5.20 x10^-7) = 0.916596 x 10^6 m/s
v = 916.6 km/s
b)
1/2*m*v^2 = h*c/lambda
lambda = 2hc/mv^2 = 2*6.626 x 10^-34 * 3 x 10^8 / 9.1 x 10^-31 * 6.6 x 10^6 * 6.6 x 10^6 = 0.100 x 10^-7 = 10.0 nm=1.0 x 10^-8 m
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