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

Answer:

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

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;

$K.E = \frac{1}{2}m_ev^2$

Apply work - energy theorem;

K.E = W

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

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