Question

An electron is accelerated using Coolidge tube of high voltage 30 KV, calculate the frequency and wave length of the produced X-ray if the electron loses one third of its velocity.

Answer 1

In order to solve this problem we need the following equations:

$E=qV$$E=\frac{hc}{\lambda}$

So:

$\begin{gathered} q=1.6\times10^{-19}C \\ E=1.6\times10^{-19}\cdot30000 \\ E=4.8\times10^{-15}J \end{gathered}$

So:

$\begin{gathered} 4.8\times10^{-15}J=\frac{hc}{\lambda} \\ _{\text{ }}where\colon \\ c=\frac{1}{3}(3\times10^8\frac{m}{s}) \\ h=6.626\times10^{-34}\cdot J\cdot s \\ _{\text{ }}solve_{\text{ }}for_{\text{ }}\lambda\colon \\ \lambda=\frac{\frac{1}{3}(3\times10^8\frac{m}{s})\cdot6.626\times10^{-34}\cdot J\cdot s}{4.8\times10^{-15}J} \\ \lambda=1.38\times10^{-11}m=0.0138nm \end{gathered}$