Question

Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative electrode), where they are produced, toward the screen, which also serves as the anode (positive electrode), 25.0 cm away.At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest, and ignore relativistic effects?

Answer 1

Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

V = Voltage = $2\times 10^4\ V$

The kinetic energy of the electron is

$K=\dfrac{1}{2}mv^2$

Energy is given by

$E=qV$

Balancing the energy

$qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s$

The velocity of the electrons is 83816746.4254 m/s