91Ó°ÊÓ

• Force acting on the electron is$F={\text{2×10}}^{\text{-12}}\text{ N}$
• The initial velocity is 0.93c
• The final velocity is 0.99c
• The mass of the electron is$m=9.1×{10}^{−31}\text{ k²µ}$

The time is calculated by using Impulse formula.

Impulse is the equivalent to the force as the change in momentum,

$\begin{array}{c}F×t=mv−muâ‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)\\ \text{where,}\\ F=\text{¹ó´Ç°ù³¦±ð a³¦³Ù¾±²Ô²µâ€„o²Ôâ€Ôò³ó±ð‱ð±ô±ð³¦³Ù°ù´Ç²Ô}⇒2×{10}^{−12}\text{ N}\\ \text{³Ù =°Õ¾±³¾±ð r±ð±ç³Ü¾±°ù±ð»å}\\ m=\text{³¾²¹²õ²õ o´Úâ€Ôò³ó±ð p°ù´Ç³Ù´Ç²Ô}⇒9.1×{10}^{-31}\text{ }\mathrm{kg}\\ v=\text{´Ú¾±²Ô²¹±ô v±ð±ô´Ç³¦¾±³Ù²â}⇒0.99\text{³¦â€„â¶Ä„â¶Ä„â¶Ä„}\left(\text{where},\text{ }c=\text{²õ±è±ð±ð»å o´Úâ€Ôò³ó±ð l¾±²µ³ó³Ù}⇒3×{10}^8\text{ m/²õ}\right)\\ v=0.99×3×{10}^8\\ =2.97×{10}^8\text{ m/²õ}\\ u=\text{¾±²Ô¾±³Ù¾±²¹±ô v±ð±ô´Ç³¦¾±³Ù²â}⇒0.93\text{³¦â€„}\left(\text{where},\text{ }c=\text{²õ±è±ð±ð»å o´Úâ€Ôò³ó±ð l¾±²µ³ó³Ù}⇒3×{10}^8\text{ m/²õ}\right)\\ u=0.994×3×{10}^8\\ =2.79×{10}^8\text{ m/²õ}\end{array}$

From Equation (1),

$\begin{array}{c}t=\frac{mv−mu}{F}\\ =\frac{\left(9.1×{10}^{-31}\text{ }\mathrm{kg}×2.97×{10}^8\text{ m/²õ}\right)\text{ -}\left(9.1×{10}^{-31}\text{ }\mathrm{kg}×2.79×{10}^8\text{ m/²õ}\right)}{2×{10}^{−12}\text{ N}}\\ =\frac{\left(9.1×{10}^{-31}\text{ }×2.97×{10}^8\right)\text{-}\left(9.1×{10}^{-31}×2.79×{10}^8\right)}{2×{10}^{−12}\text{ }}\text{ }\left(\frac{1\text{ }\mathrm{kg}â‹\dots 1\text{ m }}{1\text{ s}}−\frac{1\text{ }\mathrm{kg}â‹\dots 1\text{ m }}{1\text{ s}}\right)/1\text{ N}\\ =\frac{\left(9.1×{10}^{-31}\text{ }×2.97×{10}^8\right)\text{-}\left(9.1×{10}^{-31}×2.79×{10}^8\right)}{2×{10}^{−12}\text{ }}\text{ â¶Ä„}\left(\frac{1\text{ }\mathrm{kg}â‹\dots 1\text{ m }}{1\text{ s}â‹\dots \text{ }1\text{ N}}\right)\\ =\frac{\left(9.1×{10}^{-31}\text{ }×2.97×{10}^8\right)\text{-}\left(9.1×{10}^{-31}×2.79×{10}^8\right)}{2×{10}^{−12}\text{ }}\text{ â¶Ä„}\left(\frac{1\text{ }\mathrm{kg}â‹\dots 1\text{ m }â‹\dots 1{\text{ s}}^2}{1\text{ s}â‹\dots \text{ }1\text{ }\mathrm{kg}â‹\dots 1\text{ m}}\right)\\ =8.19×{10}^{−12}\text{ s}\end{array}$

Hence, Time is required to increase the electron's speed from 0.93c to 0.99c is $\mathbf{8}\mathbf{.19}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{12}}\mathbf{\text{ s}}$

To find the distance,

$\begin{array}{c}{x}_f=x_i+v_{xi}t\\ =0+0.93×8.19×{10}^{−12}\\ =7.62×{10}^{−12}\text{ m}\end{array}$

Hence, the electron goes in this time is $\mathbf{7}\mathbf{.62}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{12}}\mathbf{\text{ m}}$