Question

In an old-fashioned television tube, an electron () starting from rest experiences a force of 4.0 × 10–15 N over a distance of 50 cm. Ignoring the relativistic effects, the final speed of the electron is:

Answer 1

The final speed of the electron is 6.626 x 10⁷ m/s

Step-by-step explanation:

force applied to the electron, f = 4 x 10⁻¹⁵ N

distance traveled by the electron, d = 50 cm = 0.5 m

The work done on the electron;

W = F x d

W = (4 x 10⁻¹⁵ N) x (0.5 m)

W = 2 x 10⁻¹⁵ J

The kinetic energy of the electron is given by;

`K.E = (1)/(2)m_ev^2`

Apply work - energy theorem;

K.E = W

`(1)/(2)m_ev^2 = 2 *10^(-15)\\\\ v^2 = (2( 2 *10^(-15)))/(m_e)\\\\v= \sqrt{(2( 2 *10^(-15)))/(m_e)}\\\\ v = \sqrt{(2( 2 *10^(-15)))/(9.11*10^(-31))}\\\\v = 6.626*10^7 \ m/s`

Therefore, the final speed of the electron is 6.626 x 10⁷ m/s