Question

The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen

Answer 1

Answer:

The speed is  $v =10.27 *10^{7} \ m/s$

Explanation:

From the question we are told that

      The  voltage  is  $V = 30 kV = 30*10^{3} V$

      The  initial velocity of the electron is  $u = 0 \ m/s$

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      $PE = KE$

Where  

      $KE = \frac{1}{2} * m* v^2$

Here  m is the mass of the electron with a value of  $m = 9.11 *10^{-31} \ kg$

     and  

         $PE = e * V$

      Here  e is the charge on the electron with a value  $e = 1.60 *10^{-19} \ C$

=>    $e * V = \frac{1}{2} * m * v^2$

=>      $v = \sqrt{ \frac{2 * e * V}{m} }$

substituting values  

           $v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }$

          $v =10.27 *10^{7} \ m/s$