A photon scatters off an electron and the elecron\'s KE is 100keV and its veloci

Question

A photon scatters off an electron and the elecron's KE is 100keV and its velocity makes an angle of ('phi' = 37 degrees) from the direction of the original photon.

a) What is the wavelength of the original photon?

b) What is the angle, ('theta'), through which the photon is scattered?

A photon scatters off an electron and the elecron's KE is 100keV and its velocity makes an angle of ('phi' = 37 degree s) from the direction of the original photon. What is the wavelength of the original photon? What is the angle, ('theta'), through which the photon is scattered?

Explanation / Answer

kinetic energy of electron is K = 100 keV = 100 x 10^3 eV = 100 x 10^3 x 1.6 x 10^-19 J

let m be the mass of electron and v be its velocity

we know that

(1/2)mv^2 = K

or v^2 = (2K/m)

or v = (2K/m)^1/2

the wavelength of the original photon is

lamda = (h/mv)

where h = 6.63 x 10^-34 J-s

let theta be the angle through which the photon is scattered

we know that

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

where c = 3 x 10^8 m/s

solving the above equation for theta

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