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=(hc)/(\lambda)`

So:

`\begin{gathered} q=1.6*10^(-19)C \\ E=1.6*10^(-19)\cdot30000 \\ E=4.8*10^(-15)J \end{gathered}`

So:

`\begin{gathered} 4.8*10^(-15)J=(hc)/(\lambda) \\ _{\text{ }}where\colon \\ c=(1)/(3)(3*10^8(m)/(s)) \\ h=6.626*10^(-34)\cdot J\cdot s \\ _{\text{ }}solve_{\text{ }}for_{\text{ }}\lambda\colon \\ \lambda=((1)/(3)(3*10^8(m)/(s))\cdot6.626*10^(-34)\cdot J\cdot s)/(4.8*10^(-15)J) \\ \lambda=1.38*10^(-11)m=0.0138nm \end{gathered}`