Question

Separation between two parallel plates facing each other is $2 \,cm$ and surface area $l^{2}=100 \,cm ^{2} .$ If $10^{6}$ electrons of velocity $10^{8}\, m / s$ projected into the gap between plates of potential difference $400\, V$, the deflection of an electron is

• A. 17.6 mm
• B. 1.76 mm
• C. 0.176 mm
• D. zero

Answer 1

Answer: (B)

The reflection of electron is given by
$y=0+\frac{1}{2} a t^{2} $
$=\frac{1}{2}\left(\frac{e E}{m}\right)\left(\frac{l}{v}\right)^{2} $
$=\frac{e E l^{2}}{2 m v^{2}}=\frac{e V l^{2}}{2 m d v^{2}}$
$=\frac{1.6 \times 10^{-19} \times 400 \times 10^{-2}}{2 \times 9.1 \times 10^{-31} \times 2 \times 10^{-2} \times\left(10^{8}\right)^{2}}$
$=0.176 \times 10^{-2} \,m$
$=0.176 \,cm =1.76\, mm$