Question

An electron moving with the speed 5 x 106 m/s is
projected parallel to the electric field of intensity
1 x 103 N/C. Field is responsible for the retardation
of motion of electron. Find the distance travelled by
the electron before coming to rest for an instant (m.
= 9 x 10-31 kg, e = 1.6 10-19 C)
(1) 7 m
(2) 0.7 mm
(3) 7 cm
(4) 0.7 cm​

Answer 1

Given:

An electron moving with the speed 5 x 106 m/s is

projected parallel to the electric field of intensity

1 x 103 N/C. Field is responsible for the retardation of motion of electron.

To find:

Distance travelled by electron before coming to rest.

Calculation:

First, we will calculate the retardation experienced by the electron.

$\therefore \: acc. = \dfrac{force}{mass}$

$= > \: acc. = \dfrac{charge \times field \: intensity}{mass}$

$= > \: acc. = \dfrac{1.6 \times {10}^{ - 19} \times {10}^{3} }{9 \times {10}^{ - 31} }$

$= > \: acc. = 1.7 \times {10}^{14} \: m {s}^{ - 2}$

Applying 3rd Equation of Kinematics:

${v}^{2} = {u}^{2} + 2as$

$= > {0}^{2} = {(5 \times {10}^{6}) }^{2} - 2(1.7 \times {10}^{14} )s$

$= > 2(1.7 \times {10}^{14} )s = 25 \times {10}^{12}$

$= > 3.4\times {10}^{14} s = 25 \times {10}^{12}$

$= > \: s = 7.35 \times {10}^{ - 2} \: m$

$= > \: s = 7.35 \: cm$

$= > \: s \approx 7 \: cm$

So, final answer is :

$\boxed{ \red{ \bold{ \large{\: s \approx 7 \: cm}}}}$