A small piece of cesium metal ( φ = 1.9 eV ) is positioned 20 cm...

The energy of the incident photon is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{400 \text{ nm}} = 3.1 \text{ eV}$

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron is:

K_\max = E - \phi = 3.1 \text{ eV} - 1.9 \text{ eV} = 1.2 \text{ eV}

The fastest electron emitted parallel to the plate has its entire kinetic energy in the direction parallel to the plate. Its initial velocity $v$ is:

v = \sqrt{\dfrac{2 K_\max}{m}}

For a large metal plate, the electric field just outside the surface carrying a local surface charge density $\sigma$ is given by Gauss's law as:

$E_{\text{field}} = \dfrac{\sigma}{\varepsilon_0} = \dfrac{1.0 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 113 \text{ V/m}$

The electric force on the electron is perpendicular to its initial velocity, causing an acceleration towards the plate:

$a = \dfrac{e E_{\text{field}}}{m}$

The time $t$ taken for the electron to travel the distance $d = 0.2 \text{ m}$ to the plate is:

$d = \dfrac{1}{2} a t^2 \Rightarrow t = \sqrt{\dfrac{2d}{a}} = \sqrt{\dfrac{2d m}{e E_{\text{field}}}}$

The displacement $x$ parallel to the initial velocity right before it hits the plate is:

x = v t = \sqrt{\dfrac{2 K_\max}{m}} \sqrt{\dfrac{2d m}{e E_{\text{field}}}} = \sqrt{\dfrac{4 d K_\max}{e E_{\text{field}}}}

Substituting the values (note that K_\max = 1.2 \text{ e} Joules):

$x = \sqrt{\dfrac{4 \times 0.2 \text{ m} \times 1.2 \text{ eV}}{e \times 113 \text{ V/m}}} = \sqrt{\dfrac{0.96}{113} \text{ m}^2}$

$x = \sqrt{0.008496} \text{ m} \approx 0.092 \text{ m} = 9.2 \text{ cm}$