
An electron is released from the bottom plate $A$ as shown in figure $(E = {10^4}N.{C^{ - 1}})$ . The velocity of the electron when it reaches plate $B$ will be nearly equal to
A. $0.85 \times {10^7}\,m.{s^{ - 1}}$
B. $1.0 \times {10^7}\,m.{s^{ - 1}}$
C. $1.25 \times {10^7}\,m.{s^{ - 1}}$
D. $1.65 \times {10^7}\,m.{s^{ - 1}}$
Answer
491.7k+ views
Hint: In order to this question, first we will explain the reason of why there is the existence of velocity and we will write the force acts by an electron to reach toward plate B. And then we will apply the Newton’s second law to find the equation of velocity and acceleration. And then at last we will apply the formula to find the final velocity in terms of both the velocities and the acceleration.
Complete step by step answer:
As we can see in the given figure, an electron is released from the bottom plate A and it reaches plate B, that means the force is acted by an electron. So, the velocity will change in the case, as force acts.
Initial velocity is, $u = 0$.
Distance covered by an electron, $s = 2\,cm$.
So, the force of an electron, $F = q\overrightarrow E $.
Now, according to the Newton’s 2nd Law:-
$Ma = qE$
$ \Rightarrow a = \dfrac{{qE}}{M}$
where, $M$ is the mass of an electron and $a$ is the acceleration of the electron.
Now, we will apply the formula in terms of both the velocities and the acceleration:-
${v^2} - {u^2} = 2as \\
\Rightarrow {v^2} - {0^2} = 2(\dfrac{{qE}}{M})s \\ $
As we know, the mass of an electron is, $M = 9.1 \times {10^{ - 31}}Kg$
$\Rightarrow {v^2} = \dfrac{{2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^4} \times 2 \times {{10}^{ - 2}}}}{{9.1 \times {{10}^{ - 31}}}} \\
\Rightarrow {v^2} = 0.703 \times {10^{14}} \\
\therefore v = 0.83 \times {10^7} $
So, the required velocity of the electron when it reaches plate B will be nearly equal to $0.85 \times {10^7}\,m.{s^{ - 1}}$ .
Hence, the correct option is A.
Note: Subatomic particles like electrons move in random directions all the time. When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity. Due to an electric field, charged particles (e.g. electrons) in a material achieve an average velocity.
Complete step by step answer:
As we can see in the given figure, an electron is released from the bottom plate A and it reaches plate B, that means the force is acted by an electron. So, the velocity will change in the case, as force acts.
Initial velocity is, $u = 0$.
Distance covered by an electron, $s = 2\,cm$.
So, the force of an electron, $F = q\overrightarrow E $.
Now, according to the Newton’s 2nd Law:-
$Ma = qE$
$ \Rightarrow a = \dfrac{{qE}}{M}$
where, $M$ is the mass of an electron and $a$ is the acceleration of the electron.
Now, we will apply the formula in terms of both the velocities and the acceleration:-
${v^2} - {u^2} = 2as \\
\Rightarrow {v^2} - {0^2} = 2(\dfrac{{qE}}{M})s \\ $
As we know, the mass of an electron is, $M = 9.1 \times {10^{ - 31}}Kg$
$\Rightarrow {v^2} = \dfrac{{2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^4} \times 2 \times {{10}^{ - 2}}}}{{9.1 \times {{10}^{ - 31}}}} \\
\Rightarrow {v^2} = 0.703 \times {10^{14}} \\
\therefore v = 0.83 \times {10^7} $
So, the required velocity of the electron when it reaches plate B will be nearly equal to $0.85 \times {10^7}\,m.{s^{ - 1}}$ .
Hence, the correct option is A.
Note: Subatomic particles like electrons move in random directions all the time. When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity. Due to an electric field, charged particles (e.g. electrons) in a material achieve an average velocity.
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