Question

A beam of electrons of velocity $3 \times 10^{7} ms ^{-1}$ is deflected $1.5\, mm$ is passing $10\, cm$ through an electric field of $1800\, Vm ^{-1}$ perpendicular to their path. The value of $\frac{e}{m}$ for electron is

• A. $1.78 \times 10^{11} C kg ^{-1}$
• B. $2 \times 10^{11} Ckg ^{-1}$
• C. $1.5 \times 10^{11} C kg ^{-1}$
• D. $3.5 \times 10^{11} C kg ^{-1}$

Answer 1

Answer: (C)

As, $y=\frac{1}{2} a t^{2} =\frac{1}{2} \frac{e E}{m} \frac{x^{2}}{v^{2}}$
or $\frac{e}{m} =\frac{2 y v^{2}}{E x^{2}}$
$=\frac{2 \times 1.5 \times 10^{-3} \times\left(3 \times 10^{7}\right)^{2}}{1800 \times(0.1)^{2}} $
$=1.5 \times 10^{11} Ckg ^{-1}$