Question

An electron is emitted with a velocity of 5×10 6 m/s. It is accelerated by an electric field in the direction of initial velocity at 3×10 14 m/s2. If its final velocity is 7×10 6 m/s; calculate the distance covered by electron.

Answer 1

Answer : distance, s = 0.04 m

Explanation :

It is given that,

Initial velocity of electron, $u=5\times 10^6\ m/s$

Final velocity of electron, $v=7\times 10^6\ m/s$

Acceleration of electrons, $a=3\times 10^{14}\ m/s^2$

We know from the third equation of motion :

$v^2-u^2=2\ a\ s$

where,

s is the displacement of the electron.

$(7\times 10^6\ m/s)^2-(5\times 10^6\ m/s)^2=2\times 3\times 10^{14}\ m/s^2\times s$

On solving,

$s=4\times 10^{-2}\ m$

or

$s=0.04\ m$

Hence, the distance covered by the electron is 0.04 m.

Answer 2

Answer:

0.02

Explanation:

the above procedure is correct