deletions | additions       diff --git a/The_discrete_energy_levels_can__.tex b/The_discrete_energy_levels_can__.tex  index 28c5e38..ef70a2c 100644  --- a/The_discrete_energy_levels_can__.tex  +++ b/The_discrete_energy_levels_can__.tex 
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[The discrete energy levels can be explained by elastic and inelastic collisions between the electrons and the neon or argon atoms. When the accelerating voltage applied to the cathode is above a certain point, the Franck-Hertz experiment found that there are no longer only elastic collisions. The inelastic collisions occur once the electron has enough energy to excite the atom.] <-- should i remove? is this repeated? The distance the electron travels before colliding with an atom and exciting it is called the mean free path.Electrons gain additional accelerating energy as the number of inelastic collision is increasing, which can explain why the spacing between two anode voltage minima is increasing.( I DON'T KNOW WHY THE SPACING IS INCREASING IN THIS WAY).  The mean free path can be calculated using the equation below: $$\lambda=\frac{L}{2E_{a}}\frac{d\Delta E(n)}{dn}$$  where $\lambda$ is the mean free path, $L$ is the distance between accelerating grid and control grid, $\Delta E(n)$ is the spacing between two minima in Franck Hertz experiment, $n$ is the minimum order and $E_{a}$ is the lowest excitation energy.  As the applied accelerating voltage increases, electrons continue to gain energy along their mean free path and could possibility also excite the higher energy levels of the atoms. Since electrons gain additional accelerating energy as the number of inelastic collision increase, this means that if an electron inelastically collides twice with atoms their total energy gained is $$E_2=2E_a+2\delta_2$$. For n inelastic collisions, the total energy gained by an electron can be expressed as below:   $$E_n=n(E_a+\delta_n)$$  where $$\delta_n=n\frac{n}{L}E_a$$.   The two equations above can be used to derive an expression for the spacing between two minimas (refer to Appendix A). The expression, $$\Delta E(n)=[1+\frac{n}{L}(2n-1)]E_a$$, shows that the spacing between two anode voltage minima is increasing.  In this lab, Ne atoms have only been excited to the first and  second energy level and level; whereas,  Ar atoms mainly stay in the first energy level. According to the NIST Atomic Spectra Database, the first excitation energy of Ne I is 16.62eV, and the second excitation energy level is 18.38eV. For Ar I, the four sub levels within the first energy level are 11.55eV, 11.62eV, 11.72eV and 11.83eV respectively.(INCLUDE DIAGRAM)    More stuff to discuss...  \begin{itemize}