Question

Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.

Answer 1

Explanation:  

To Find: The displacement of this electron parallel to its initial velocity before it strikes the large metal plate.

Metal plate electric field,

$E=\frac{\sigma}{\varepsilon_{0}}=\frac{1 \times 10^{-9}}{8.85 \times 10^{-12}}$

           $\begin{aligned}&=\frac{1 \times 10^{3}}{8.85}\\&=113 \mathrm{V} / \mathrm{m}\end{aligned}$

Acceleration, $a=\frac{q E}{m}$

Where, q = electron charge          

             E = electric area          

            M = electron  mass  

     $\begin{aligned}&a=\frac{1.6 \times 10^{-19} \times 113}{9.1 \times 10^{-31}}\\&a=\frac{1.6 \times 10^{12} \times 113}{9.1}\end{aligned}$

     $\begin{aligned}a &=\frac{180.8 \times 10^{12}}{9.1} \\&=19.87 \times 10^{12}\end{aligned}$

      $\begin{aligned}t &=\sqrt{\frac{2 y}{a}}=\sqrt{\frac{2 \times 20 \times 10^{-2}}{19.87 \times 10^{12}}} \\&=1.41 \times 10^{-7} \mathrm{s}\end{aligned}$

From the Photoelectric Equation of Einstein,

$K . E .=\frac{h c}{\lambda}-W=1.2 \mathrm{eV}$

         $=1.2 \times 1.6 \times 10^{-19} \mathrm{J}$

         [∵ K.E. = 1.2 eV]

∴Velocity,

      $v=\sqrt{\frac{2 K E}{m}}$

         $=\sqrt{\frac{2 \times 1.2 \times 1.6 \times 10^{-19}}{4.1 \times 10^{-31}}}$

         $=0.665 \times 10^{-6} \mathrm{m} / \mathrm{s}$

∴ Horizontal displacement,  

S = v × t

$S=0.665 \times 10^{-6} \times 1.4 \times 10^{-7} \mathrm{S}$

   = 0.092 m

   = 9.2 cm