Question

An electron initially at rest recoils after a head-on collision with a 8.27-keV photon. Determine the kinetic energy acquired by the electron.

Answer 1

we know, energy of photon, E₁ = h*c/lamda₁

lamda₁ = h*c/E₁

lamda₁ = 6.625*10^-34*3*10^8/(8.27*10^3*1.6*10^-19)

lamda₁ = 1.5*10^-10 m

after the collision, the photon come back in the same direction.

using compton effect equation,

the change in wavelength of photon,

lamda₂ - lamda₁ = (h/(m_o*c))*(1 - cos(180))

= 2*h/(m_o*c)

lamda₂ = lamda₁ + 2h/(m_o*c)

= 1.5*10^-10 m + 2*6.625*10^-34/(9.1*10^-31*3*10^8)

lamda₂ = 1.5458*10^-10 m

Energy of recoiled photon, E₂ = h*c/lamda₂

E₂ = 6.625*10^-34*3*10^8/(1.5458*10^-10)

= 1.29*10^-15 J

= (1.29*10^-15)/(1.6*10^-19) eV

E₂ = 8.036 keV

now Apply conservation of energy

gain in kinetic energy of electron = loss electrons kinetic energy

= E₁ - E₂

= 8.27 keV - 8.036 keV

kinetic energy acquired by the electron = 234 eV