A cathode ray tube uses a potential difference of 5.0 kV to accelerate electrons and produce an electron beam that makes images on a phosphor screen. What is the speed of these electrons as a percentage of the speed of light? multiple choice0.025%0.22%1.3%4.5%14%25%

calculate the velocity of the electron as a percentage of the speed of light.round your answer to the first decimal place.

The diagram below represents a cathode ray tube in which electrons emitted from the hot filament are accelerated towards the positively charged screen. The electric field in the evacuated space between the screen and filament has a constant magnitude of 105 Vm-1. The speed of the electrons on reaching the screen is:Select one:a.8.4x107 ms-1.b.1.8x1016 ms-1.c.2.0x103 ms-1.d.1.0x104 ms-1.e.3.7x106 ms-1.Clear my choice

Electrons emitted from the cathode of a cathode ray tube are accelerated by a potentialdifference of 2.00 kV between the cathode and the anode. The electrons then enter the uniformelectric field between the middle of two parallel conducting plates, each of length 2.00 cm. ThePotential difference between the plates is 500 V and the plates are separated by a distance of1.50 cm.Calculate(a) the velocity 𝑢 of the electrons when they emerge from the anode.(b) the time taken for the electrons to transverse the uniform electric field.(c) the velocity of the electrons when they emerge from the uniform electric field.

An electron is accelerated from rest between two parallel plates that are seperated by 2.20 cm. The potential difference between the two plates is 2.10×103 V. If the electron travels from the negative plate to the positive plate, what will be the speed of the electron as it exits through a small hole in the positive plate? 2.72×107 m/s 2.20×107 m/s 3.24×107 m/s 4.03×106 m/s

Calculate the speed achieved by an electron accelerated in a vacuum through a p.d of 2000 V. (5 marks)