Question

The following figure is a simplified diagram of a commercially available EI source.
(a) What voltage must be applied between the filament and target so that electrons interacting with molecules
at the point marked SS (sample source) will have 70 eV of kinetic energy?
(b) What will happen to a molecule that diffuses toward the filament and is ionized at point P?

Short answer

a) The voltage applied is 140 V

b) It will collide with the solid part at exit slit.

Step-by-step solution

part(a) step 1: Given Information

Given kinetic energy is 70 eV.

part(a) step 2: Explanation

$V=70\mathrm{V}$The the kinetic energy is given by:

$\mathrm{KE}=zeV........\left(1\right)$
Where,
z= charge on the exiting slit is
e= electronic charge
V= voltage difference between two slits

Now substitute the values in equation (1)

Substitute KE=70 eV , $z=1.6\times {10}^{-19}\mathrm{C}$and e=+1, we get

$70\mathrm{eV}=(+1)\left(1.6\times {10}^{-19}\mathrm{C}\right)\mathrm{V}$

$\left(70\mathrm{eV}\times \frac{1.6\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}\times \frac{1\mathrm{C}·\mathrm{V}}{1\mathrm{J}}\right)=(+1)\left(1.6\times {10}^{-19}\mathrm{C}\right)\mathrm{V}$

Simplify ,

$V=70\mathrm{V}$

This is the potential needed for moving half the distance.

So potential needed to move to full distance is $2\times 70=140V$

Part(b) Step 1 : Explanation

Electron ionization (EI), also known as electron impact ionization or bombardment method, is an ionization method in which energetic electrons interact with gas-phase atoms or molecules to produce ions.
When an ion is formed at the point P, it will collide with the solid part of the slit at the exit. This phenomenon occurs due to the Repeller accelerating plate voltage.
This causes the molecules to diffuse towards the filament to ionize at the point P.