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The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen?

Respuesta :

cjrodriguez89

Answer:

v= 8.8*10⁷ m/s

Explanation:

  • Assuming no friction present, the change in electrical potential energy, must be equal in magnitude, to the change in kinetic energy of the electron.
  • The change in the electrical potential energy, can be expressed as follows:

       [tex]\Delta U = (-e)*\Delta V[/tex]

  • The change in kinetic energy, assuming that the electron started from rest, can be written as follows:

       [tex]\Delta K = \frac{1}{2} *m*v^{2}[/tex]

       ⇒[tex]\Delta K = -\Delta U[/tex]

  • From the equation above, replacing by ΔK and ΔU, we have:

       [tex]-\Delta U =- (-e)*\Delta V =\Delta K = \frac{1}{2} *m*v^{2}[/tex]

  • Solving for v:

        [tex]v=\sqrt{\frac{2*e*\Delta V}{m_{e}} } =\sqrt{\frac{2*1.6e-19C*22e3V}{9.1e-31kg}} }\\ v= 8.8e7 m/s[/tex]

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