Answer
a) $ 5.17 \times 10^{-11} \mathrm{m} $
b) $5.24 \times 10^{-11} \mathrm{m} $ .
c) $3.7935 \times 10^{-15} \mathrm{J}$
Work Step by Step
a) We know that X rays can be produced when electrons accelerated to high kinetic energy across a potential increase $V_{AC}$ strike a target, from the following formula $$ e V_{AC} = hf $$
- $V_{AC} $ is the potential difference , $f$ is the frequancy of the emitted
x-rays , $h$ is planck constant and $e$ is the charge of the accelerated electron .
From the formula of the velocity of the photon , we can write : $$ hf = h \frac{c}{\lambda} = e V_{AC} $$
So , from given value :$ e = 1.6 \times 10^{-19} \mathrm{C}$ , $ c = 3 \times 10^{8} \mathrm{m/s}$ , $ V_{AC} = 24 \times{10^{2}}$ and $h = 6.626 \times 10{-34} \mathrm{J \cdot s}$.
- Now, substitute :
$\lambda = \dfrac{ h c}{ e V_{AC}} = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{ 1.6 \times 10^{-19} \times 24 \times 10^{3}} = 5.17 \times 10^{-11} \mathrm{m} $
b) From compton scattering relation : $ \lambda_{1} - \lambda = \dfrac{h}{mc} ( 1 - \cos \phi) $
- $ \phi $ is the scattering angle , $h $ is planck constant , $m$ is the mass of the electron , $ c$ is the velocity of the light , $\lambda_{1} $ is the wavelength of the scattered photon .
- From the known value : $ m = 9.1 \times 10^{-31} \mathrm{kg} $ and $ \phi = 45$.
- Now, substitute ,
$ \lambda_{1} = \lambda + \dfrac{h}{mc} ( 1 - \cos \phi) = 5.17 \times 10^{-11} + \dfrac{6.626 \times 10^{-34}}{ 9.1 \times 10^{-31} \times 3 \times 10^{8} } (1 - \cos 45) = 5.24 \times 10^{-11} \mathrm{m} $ .
c) Now , the energy of the scattered photon :
$ E = hf = \dfrac{hc}{ \lambda_{1}} = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^{8} }{ 5.24 \times 10^{-11}} = 3.7935 \times 10^{-15} \mathrm{J}$
