Update vol-I-chap-2-sect-1.md

docs/vol-I/vol-I-chap-2-sect-1.md

@@ -92,7 +92,7 @@
 
 	The region of **Preparation** consisted in a vacuum tube containing a negative cathode (_C_), a positive anode (_A_) and a slit (_S_): when _C_ is heated, a radiation is emitted, attracted by _A_ and collimated trough _S_. The initial conditions are the following: a particle of mass $m$ and charge $q$ leaves the cathode and arrives at the condenser with a horizontal velocity $v_{horizontal} = v_0$ in a direction parallel to the plates _D_ and _E_.
 
-	The region of **Transformation** corresponds to the inside of two charged metallic plates of a condenser (_D_ and _E_) where the external electric and magnetic fields will be applied. If the electrostatic force $F_{\mathrm{electric}} = qE$. is applied, a transversal acceleration is generated in a direction perpendicular to the incident beam $a_{\mathrm{transversal}} = a_t = qE/m$. Such force acts during the time $t = d/v_0$ required by the particle to travel the distance $d$ inside the condenser with a transversal velocity $v_{\mathrm{transversal}}= (a_t)(t)= (\frac{qE}{m})(\frac{d}{v_0})=  \frac{qEd}{mv_0}$ is acquired by the particle.       
+	The region of **Transformation** corresponds to the inside of two charged metallic plates of a condenser (_D_ and _E_) where the external electric and magnetic fields will be applied. If the electrostatic force $F_{\mathrm{electric}} = qE$. is applied, a transversal acceleration is generated in a direction perpendicular to the incident beam $a_{\mathrm{transversal}} = a_t = qE/m$. Such force acts during the time $t = d/v_0$ required by the particle to travel the distance $d$ inside the condenser with a transversal velocity $v_{\mathrm{transversal}}= (a_t)(t)= (\frac{qE}{m})(\frac{d}{v_0})=  \frac{qEd}{mv_0}$. 
 
 	Inside this region of Transformation, the displacement of the particle has two components: the horizontal component $x = v_0t$ due to the constant velocity $v_0$ and the variable vertical component $y = \frac{1}{2} (a_t)t^2$ produced by the transversal acceleration $a_t$. Taking into account these two components of the displacement, the resulting equation of the trajectory inside the condenser is a parabola $y =  \frac{1}{2} (a_t)t^2 =  \frac{1}{2} (\frac{qE}{m})[(\frac{x}{v_0})^2] =  \frac{1}{2} (\frac{qE}{m{v_0}^2})(x^2)$ which correspond to the form $y = Ax^2$.
 
@@ -165,13 +165,13 @@
 
 	What was observed in the experiment was the following: When the accelerating potential V started to increase the current, I also increased. At a critical value $V_c = 4.9$ volts the current went down sharply; however, when the voltage increased again the current increased also. In arriving at a new critical value $2V_c = 9.8$ volts, the same strong diminution of the current was observed. These observations can be interpreted as follows:
 
-	- When $V < V_c = 4.9$ volts the accelerated electrons have low kinetic energy and their collisions with the atoms are elastic ones: the mass of the electron is much smaller than the mass of the atom, therefore the incident electron rebounds and does not give energy to the electrons in the atom.
-
+	- When $V < V_c = 4.9$ volts the accelerated electrons have low kinetic energy and their collisions with the atoms were elastic ones: as the mass of the electron is much smaller than the mass of the atom, the incident electrons rebound and do not give energy to the electrons in the atom.
+	
 	- When $V = V_c$ the Mercury atom receives the exact amount of energy to excite their electrons to higher energy levels. Under this condition the collision is inelastic because the incoming electron lost the energy that gives up to the electrons inside the atom; therefore, the current I diminishes.
 
 	- If now $V > V_c$ the current I increases until $V = 2V_c$ and the collisions become inelastic again. These fluctuations in the current repeat when the voltage V attains a multiple value of $V_c$.
 
-	It is interesting to note that in the atom of Mercury the difference between the lowest energy level and the first excited energy level corresponds to a spectral line whose wavelength is $\lambda = 2536 Å$.  For photons $E = hν = hc/\lambda$, if now we make $E = qV_c$ where $q$ is the charge of the electron, we can get $\lambda = (hc)/(qV_c) = 2,536 \times 10^{-10} m = 2536 Å$. M,any implications of this experiment are considered in Hertz´s Nobel Lecture The Results of the Electron-Impact Tests in the Light of Bohr’s Theory of Atoms.
+	It is interesting to note that in the atom of Mercury the difference between the lowest energy level and the first excited energy level corresponds to a spectral line whose wavelength is $\lambda = 2536 Å$.  For photons $E = hν = hc/\lambda$, if now we make $E = qV_c$ where $q$ is the charge of the electron, we can get $\lambda = (hc)/(qV_c) = 2,536 \times 10^{-10} m = 2536 Å$. Many implications of this experiment are considered in Hertz´s Nobel Lecture The Results of the Electron-Impact Tests in the Light of Bohr’s Theory of Atoms.
 
 	More details concerning Bohr´s atomic theory will be discussed in section 6.1. *Electronic energy levels in the hydrogen atom* corresponding to chapter *6. Spectroscopical studies of atomic structures.* Also, in section 6.3 of that chapter the Bohr´s Nobel Lecture *The structure of the atom* will be considered.