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?

Short answer

(a) Using the formula of Kinetic energy, $\mathrm{KE}=zeV$,the potential is calculated. Hence, the voltage applied between filament and target is$140V$.

(b) Due of the Repeller accelerating plate voltage, the molecule diffuses toward filament and it will be ionized at point $P$.

Step-by-step solution

Part (a)

The kinetic energy $\left(\mathrm{KE}\right)$is given by the equation.

$\mathrm{KE}=zeV$

Here, $z$is charge on exiting slit, $e$is the electronic charge, and $V$is voltage difference between two slits.

The kinetic energy of SS (sample is source) is $70\mathrm{eV}$.

The electronic charge is $1.6\times {10}^{-19}\mathrm{C}$.

The charge on existing slit is $+1$.

Note that all ions have same number of charges, so $z$value is assumed to be $+1$.

Now, convert the units of $\mathrm{eV}$into coulomb-volts (C-V) using the following conversion factor.

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

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

Substitute the all known values in the above equation, we get

$\mathrm{KE}=zeV$

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

$V=70\mathrm{V}$

Thus, the potential required is $70\mathrm{V}$.

This is the potential which requires to move the electron for half of the distance. It means that the potential is applied either between filament and SS and SS and Target. So, the total voltage applied between filament and target is equal to, $70\mathrm{V}\times 2=140\mathrm{V}$.

Therefore, the voltage applied between filament and target is $140\mathrm{V}$.

Part (b)

An ion formed at point P will almost certainly collide with a solid part of the exit slit. This is because of the Repeller accelerating plate voltage. That's why the molecule diffuses toward filament and it will be ionized at point $P$.