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, E_{1} =
h*c/lamda_{1}

lamda_{1} = h*c/E_{1}

lamda_{1} =
6.625*10^-34*3*10^8/(8.27*10^3*1.6*10^-19)

lamda_{1} = 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,

lamda_{2} - lamda_{1} = (h/(m_{o}*c))*(1
- cos(180))

= 2*h/(m_{o}*c)

lamda_{2} = lamda_{1} + 2h/(m_{o}*c)

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

lamda_{2} = 1.5458*10^-10 m

Energy of recoiled photon, E_{2} =
h*c/lamda_{2}

E_{2} = 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

**E _{2} =
8.036 keV**

now Apply conservation of energy

gain in kinetic energy of electron = loss electrons kinetic energy

= E_{1} - E_{2}

= 8.27 keV - 8.036 keV

**kinetic energy
acquired by the electron = 234 eV**

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