Answer
$99.45\;pm$
Work Step by Step
X rays are produced in an x-ray tube by electrons accelerated through an electric potential difference of $V=50.0\;kV$.
The kinetic energy of the electron at the end of the acceleration is
$K_0=1.6\times 10^{-19}\times50\times10^3\;J=8\times10^{-15}\;J$
The electron collides with a target nucleus (the nucleus remains stationary) and then the kinetic energy of the electron becomes
$K_1=0.500K_0=0.500\times8\times10^{-15}\;J=4\times10^{-15}\;J$
Thereafter, the electron collides with another target nucleus (the nucleus remains stationary) and the kinetic energy of the electron after second collision becomes
$K_2=0.500K_1=0.500\times4\times10^{-15}\;J=2\times10^{-15}\;J$
Therefore, the energy $(E)$ associated with the second emitted photon is
$E=K_1-K_2$
or, $E=4\times10^{-15}-2\times10^{-15}\;J=2\times10^{-15}\;J$
$\therefore$ The wavelength associated with the second emitted photon is
$\lambda=\frac{hc}{E}$
or, $\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{2\times10^{-15}}\;m$
or, $\lambda=9.945\times10^{-11}\;m$
or, $\lambda=99.45\;pm$
