Question

An electron at rest is accelerated through a potential difference of 200V.if the specific charge of the electron is $1.76\times {{10}^{11}}Ck{{g}^{-1}}$, the velocity acquired by the electron is
$\begin{align}
  & a)8.4\times {{10}^{5}}m{{s}^{-1}} \\
 & b)8.4\times {{10}^{6}}m{{s}^{-1}} \\
 & c)4.2\times {{10}^{5}}m{{s}^{-1}} \\
 & d)4.2\times {{10}^{6}}m{{s}^{-1}} \\
\end{align}$

Answer 1

Hint: When an electron is accelerated across a potential difference, it gains energy in the form of kinetic energy. This energy is nothing but equal to the work done in moving a charge across the given potential difference. Hence we will determine the change in kinetic energy and equate it to the work done in order to obtain the velocity acquired by the electron.
Formula used:
$K.E=\dfrac{1}{2}m{{v}^{2}}J$
$W=Vq$

Complete answer:
Let us say a charge ‘q’ moves through a potential difference of ‘V’. Hence the energy gained by the charge or the work done (W) is given by,
$W=Vq$
Charge when moves through a potential difference, it continuously accelerates and hence its velocity ‘v’ keeps on changing. Let us say the mass of the charge is m, then its kinetic energy at velocity v is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}J$
In the above question it is given to us that the electron accelerates from rest i.e. its initial velocity is zero. Hence the change in kinetic energy $(\Delta K.E)$ is equal to,
$\begin{align}
  & \Delta K.E=\dfrac{1}{2}m{{v}_{FINAL}}^{2}-K.E=\dfrac{1}{2}m{{v}_{INITIAL}}^{2} \\
 & \Rightarrow \Delta K.E=\dfrac{1}{2}m{{v}_{FINAL}}^{2}-K.E=\dfrac{1}{2}m{{(0)}^{2}} \\
 & \Rightarrow \Delta K.E=\dfrac{1}{2}m{{v}_{FINAL}}^{2} \\
\end{align}$
Now since, this change kinetic energy gained by the electron is equal to the work done in moving a electron along potential difference V we can write,
$\begin{align}
  & \Delta K.E=W \\
 & \Rightarrow \dfrac{1}{2}m{{({{v}_{FINAL}})}^{2}}=Vq \\
 & \Rightarrow {{v}^{2}}=\dfrac{2Vq}{m}\text{, }\because \text{q/m = }1.76\times {{10}^{11}}Ck{{g}^{-1}} \\
 & \Rightarrow {{v}^{2}}=2\times 200\times \text{ }1.76\times {{10}^{11}} \\
 & \Rightarrow {{v}^{2}}=7.04\times {{10}^{13}} \\
 & \Rightarrow v=8.39\times {{10}^{6}}m{{s}^{-1}}=8.4\times {{10}^{6}}m{{s}^{-1}} \\
\end{align}$

Hence the correct answer of the above question is option b.

Note:
In the above question we have replaced ${{v}_{FINAL}}$ as ‘v’ for sake of simplicity. It is the velocity of the electron when it crosses the region of potential difference. If the electron is further not moving in any other region of potential then it will move with the above velocity throughout, provided there is no loss of momentum.