Question

Q2. A monochromatic X-ray beam, whose wavelength is 0.558 A,
is scattered off an electron through scattering angle of 460. Find
the wavelength of the scattered beam and the kinetic energy of
the recoil electron.​

Answer 1

Given wavelength of monochromatic X-ray beam is 0.558A°

Scattering angle = 460°

$we have to find$ ;

$From compton effect;$

$\lambda'-\lambda=\lambda_c(1-cos\theta)$

$\lambda_c$ $is compton wavelength and its value is 2.43 pm.$

$\lambda'-\lambda=2.43*(1-cos460)$

$\lambda'-\lambda=0.583pm$

$\lambda'=\lambda +.583pm$

$\lambda'=558.583pm$                                 $[ As \lambda=0.558A given]$

$As electron is a particle do from de- broglie eq^n$

$\lambda=\frac{h}{mv}$

$v=\frac{h}{\lambda'm}$

$v=1.18*1o^7 m/sec$

$\therefore Kinetic energy = \frac{1}{2} mv^2$

$\therefore Kinetic energy = 6.34*10^_-17 J$

$\therefore$The wavelength of the scattered beam $\lambda'=558.583pm$

The kinetic energy of

the recoil electron