Question

If an electron is placed in an electric field with a potential difference of (1.35x10^2) V, the maximum speed the electron can attain is?

Answer 1

`rArr v_f = sqrt((2eDeltaV)/m)`

Explanation:

Use the work-energy principle or conservation of energy.

Let
V = electric potential
U = potential energy = qV
K = kinetic energy `= 1/2mv^2`
`v_i=`initial velocity = 0
`v_f =` final velocity
e = electron charge = `1.602 xx10^-19 C`
q = charge = -e
m= mass of the electron = `9.11 xx10^-31`

Most people are familiar with conservation of energy:

`Delta(U +K) =0 rArr DeltaK= - DeltaU`

`rArr 1/2mv_f^2- 1/2mv_i^2 = -qDeltaV`
`rArr 1/2mv_f^2 - 0 = -(-e)DeltaV`
`rArr v_f = sqrt((2eDeltaV)/m)`

By estimation,
`v_f ~~10^7 m/s`

Answer 2

`6.89*10^6ms^(-1)`

Explanation:

Here the electron will gain kinetic energy while losing electric potential energy. From this we can get:
`Delta(1/2mv^2)=Delta(QV)`

We can rearrange that equation:
`mv^2=2eV`
`v^2=(2eV)/m`
`v=sqrt((2eV)/m`

We know that:
`e=1.6*10^(-19)color(white)(l)C`
`V=1.35*10^2color(white)(l)V`
`m=9.11*10^(-31)kg`
`v=sqrt((2(1.6*10^(-19))(1.35*10^2))/(9.11*10^(-31)))~~6886248.41~~6.89*10^6ms^(-1)`