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

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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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

electrostatic potential energy is given as

now by energy conservation

given that for electron

now by plug in values

So here it is accelerated through potential difference of 2559.4 Volts

photocells convert light energy to electricity

think its the third option