Question

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?

Answer 1

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:

       $\Delta U = (-e)*\Delta V$

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

       $\Delta K = \frac{1}{2} *m*v^{2}$

       ⇒$\Delta K = -\Delta U$

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

       $-\Delta U =- (-e)*\Delta V =\Delta K = \frac{1}{2} *m*v^{2}$

• Solving for v:

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