Question

Assume that the electrons in a transmission electron microscope are accelerated through a potential of 80,000 volts. Determine (a) The electron velocity given by this potential by assuming that the energy the electrons gain falling through the potential equals the gain in their kinetic energy.

Answer 1

To solve this problem we will apply the concepts related to energy conservation. We will start by defining the expressions of the electric potential energy for a given charge (and for the electron). With this we can apply the conservation of kinematic energy. Our values are given as

$V = 80000V$

The potential energy:

$U = qV$

Here,

q  = Charge

V = Voltage

Or specifically for an electron we can define it as,

$U_e = eV$

Here,

e = Charge of electron

V = Voltage

Applying the energy conservation equations we have that the kinetic energy must be equivalent to the electric potential energy,

$KE= U_e$

$\frac{1}{2}mv^2 = qV =e V$

Here

v = Velocity

m = Mass

Rearranging,

$v^2 = \frac{2eV}{m}$

Replacing,

$v^2 = \frac{2(1.6*10^{-19})(80000)}{9.1*10^{-31}}$

$v= 2.81*10^{16}$

For each electron the velocity is,

$v = 1.68*10^8m/s$

Therefore the velocity of the electron is $1.68*10^8m/s$