Question

A stream of electrons enters an electrical field normal to the lines of force with a velocity of $3\times {10}^{7}m/s$. The electric intensity is $1800v/m$. While traveling through a distance of $10$ $cm$, the electron beam is deflected by $2$ $mm$. Then $e/m$ value in coulomb per kg is

• A. $2\times {10}^{10}$
• B. $2\times {10}^7$
• C. $2\times {10}^{11}$
• D. $2\times {10}^4$

Answer 1

Answer: (C)

$2\times {10}^{11}$

Force exerted by the electric field on the electrons $=1800\times e$

Acceleration experienced by electrons $=\frac{1800e}{m}$
Time taken to move 0.1m

$=\frac{0.1}{3\times {10}^7}S$

$=3.333\times {10}^{-9}$
In the above electrons deflected 2mm,

from $[S=\frac{1}{2}a{t}^2]2\times {10}^{-3}=\frac{1}{2}\times \frac{1800e}{m}\times (3.333\times {10}^{-9}{)}^2$

$\Rightarrow \frac{4\times {10}^{-3}\times {10}^{18}}{1800\times (3.3333{)}^2}=\frac{e}{m}$

$\Rightarrow \frac{e}{m}=2.0000\times {10}^{11}$

So, $\frac{e}{m}=2\times {10}^{11}$

So, the answer is option (C).