Question

Across what potential difference does an electron have to be accelerated in order to reach the speed v=3e7 m/s?

Answer 1

When an electron is accelerated through potential difference then the speed that it attain will be explained by energy conservation

here by energy conservation we can say that

change in kinetic energy of electron = electrostatic potential energy gained through given potential difference

kinetic energy is given as

$KE = \frac{1}{2}mv^2$

electrostatic potential energy is given as

$PE = qV$

now by energy conservation

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

given that for electron

$m = 9.1 * 10^{-31} kg$

$v = 3 * 10^7 m/s$

$q = 1.6 * 10^{-19} C$

now by plug in values

$1.6 * 10^{-19} * V = \frac{1}{2}*9.1* 10^{-31} *(3*10^7)^2$

$V = \frac{4.095 * 10^{-16}}{1.6 * 10^{-19}}$

$V = 2559.4 Volts$

So here it is accelerated through potential difference of 2559.4 Volts

Answer 2

The electric potential is 4100 V or 4.1kV.

We know that the kinetic energy of the moving electron is equal to the 1/2mv^2. On this basis, we can write;

1/2mv^2 = eV

Where;

m = mass of the electron

v = speed of the electron

V = potential of the electron

e = charge on the electron.

Hence;

V = 1/2mv^2 /e

V = 0.5 × 9.11 × 10^-31 × (3 × 10^7)^2/1.6 ×10^-19

V = 4100 V or 4.1kV

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