Question

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

• A. $2\times {10}^{-11}$
• B. $2\times {10}^7$
• C. $2\times {10}^{11}$
• D. $2\times {10}^{14}$

Answer 1

Answer: (C)

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

Time during which the electron stream travels 10 cm , $t=\frac{10\times {10}^{-2}}{3\times {10}^7}=\frac{1}{3}\times {10}^{-8}$

Vertical displacement during this time $=2mm$

$\frac{1}{2}a\left(t^2\right)=2\times {10}^{-3}$

$a=\frac{2\times {10}^{-3}\times 2}{\frac{1}{9}\times {10}^{-16}}=36\times {10}^{13}$

$\frac{eE}{m}=a=36\times {10}^{13}$

$\frac{e}{m}=\frac{36\times {10}^{13}}{1800}=2\times {10}^{11}$

So, the answer is option (C).