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.

(Use h = 6.63 × 10^-34J-s = 4.14 × 10^-15 eV-s, c = 3 × 10⁸ m/s and me = 9.1 × 10^-31kg)

Answer 1

Electric field of the metal plate,

`E = σ/∈_0 = (1 xx 10^-9)/(8.85 xx 10^-12)`

= 113 V/m

Acceleration,

`a = (qE)/m`,

where q=charge on electron           

          E=electric field           

          m=mass of electron

`a = (1.6 xx 10^-19 xx 113)/(9.1 xx 10^-31) = 19.87 xx 10^12`

`t = sqrt((2y)/a) = sqrt((2 xx 20 xx 10^-2)/(19.87 xx 10^12)`

=`1.41 xx 10^-7` s

From Einstein's photoelectric equation,

`K.E. = (hc)/lambda - W = 1.2  "eV"`

= `1.2 xx 1.6 xx 10^-19  "J"...........[because "in problem " 31 : "KE" = 1.2  "eV"`]

`therefore "Velocity", v = sqrt({2KE)/m)`

`= sqrt((2 xx 1.2 xx 1.6 xx 10^-19)/(4.1 xx 10^-31))` `sqrt((2 xx 1.2 xx 1.6 xx 10^-19)/(4.1 xx 10^-31))`

`= 0.665 xx 10^-6  "m/s"`

∴ Horizontal displacement,

`S = v xx t`

`S = 0.665 xx 10^-6 xx 1.4 xx 10^-7`

`S = 0.092  "m" = 9.2  "cm"`