Question

An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions of phosphorus, each of mass mP = 5.18 × 10-26 kg. The phosphorus ions are doubly ionized, with each phosphorus ion lacking two electrons. The ions start at rest at one end of the vacuum chamber and are accelerated by an electric field over a distance of re = 46 cm before they strike the silicon layer with velocity vP = 180 m/s. Calcultae the average electric field strength, E, across the vacum chamber.

Answer 1

Answer:

$E=5.7\times 10^{-3}\ V/m$

Explanation:

Given that

$m_p=5.18\times 10^{-26}\ kg$

re= 46 cm

Vp= 180 m/s

We know that

$E=\dfrac{\Delta V}{r}$

$\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}$

So

$E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}$

Now by putting the all given values in the questions

$E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}$

$E=5.7\times 10^{-3}\ V/m$

So the average electric field is $E=5.7\times 10^{-3}\ V/m$.