Question

The potential difference from the cathode (negative electrode) to the screen of an old television set is +22,000 V. An electron leaves the cathode with an initial speed of zero.

a)Determine the kinetic energy of the electron.

b)Determine the speed of the electron.

Answer 1

(a) the kinetic energy of the electron is given as ::

using a condition, P.E = - K.E

P.E = q V                                         { eq.1 }

where, V = potential difference = +22 x 10³ V

q = charge on electron = -1.6 x 10^-19 C

inserting the values in eq.1,

P.E = (-1.6 x 10^-19 C) (+22 x 10³ V)

P.E = - 35.2 x 10^-16 J

P.E = - 3.52 x 10^-15 J

which equals to kinetic energy, K.E = 3.52 x 10^-15 J

(b) the speed of the electron is given as ::

K.E = (0.5) m_e v²                                     { eq. 2 }

where, m_e = mass of the electron = 9.1 x 10^-31 kg

inserting the values in eq.2,

(3.52 x 10^-15 J) = (0.5) (9.1 x 10^-31 kg) v²

v² = (3.52 x 10^-15 J) / (4.55 x 10^-31 kg)

v = 0.773 x 10¹⁶ m/s

v = 7.73 x 10¹⁵ m/s

v = 8.79 x 10⁷ m/s