Respuesta :
Find the kinetic energy of the electron after moving through the potential difference of 702.06kV.
The electric potential energy of a charge is transformed in kinetic energy when it passes through a potential difference V. Then, the kinetic energy of the electron is:
[tex]K=e\cdot V[/tex]Where e is the charge of an electron:
[tex]e=1.602\times10^{-19}C[/tex]On the other hand, the kinetic energy of a particle with mass m is related to its speed:
[tex]\frac{1}{2}mv^2=K[/tex]In this case, m is the mass of the electron:
[tex]m=9.11\times10^{-31}kg[/tex]Then:
[tex]\frac{1}{2}mv^2=eV[/tex]Isolate the speed v from the equation:
[tex]v=\sqrt{\frac{2eV}{m}}[/tex]Replace the values for e, m and V=702.06kV to find the speed of the electron:
[tex]v=\sqrt{\frac{2(1.602\times10^{-19}C)(702.06\times10^3V)}{9.11\times10^{-31}kg}}=4.969\times10^8\frac{m}{s}[/tex]Notice that the speed of the electron calculated using the classical method is greater than the speed of light:
[tex]c=3\times10^8\frac{m}{s}[/tex]Then, the theory of Special Relativity is required to find a proper answer for this problem.
Therefore, the (inadequate) answer is: the electron is moving at a speed of approximately 4.97*10^8m/s according to Classical Physics.
