In a Compton scattering experiment, X-rays are detected at a scattering angle of

Question

In a Compton scattering experiment, X-rays are detected at a scattering angle of theta = 158 degree with a wavelength of 0.1747 nm. (1 nm = 1 times 10^-9 m; mass of electron: 9.11 times 10^-31 kg) A) Find the wavelength of an incident photon. B) What is the energy of an incident photon? C) If the scattered X-ray has energy of 1.065 times 10^-15 J, what is the kinetic energy of the recoil electron? A singly ionized helium atom He^+ (Z = 2) contains one electron in the ground state. Determine the maximum wavelength of incident radiation that can be used to remove this electron from the atom.

Explanation / Answer

theta = 158 degrees

lamda = 0.1747 nm

(a) from comphton effect

lamda - lamdao = (h/mc)(1-cos(theta))

0.1747*10^-9 -lamdao = (6.63*10^-34/(9.1*10^-31*3*10^8))(1-cos(158))

lamdao = 0.17002 nm

(b) E = hc/lamdao

E = 6.63*10^-34*3*10^8/(0.17002*10^-9)

E = 1.17*10^-15 J

(c) KE = E - Es

= 1.17*10^-15 - 1.065*10^-15

KE= 1.05*10^-16 J

2) E = -13.6z^2/n^2

z = 2 , n =1

E = -13.6*2^2/1^2

E = -54.4 eV

E(eV) =1243/lamda(nm)

54.4 = 1243/lamda

lamda = 22.85 nm

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