A metal plate with a work function of 2.2 eV is illuminated by a...

Energy of the incident photon is given by:

$E = \dfrac{hc}{\lambda}$

Using $hc = 1240 \text{ eV}\cdot\text{nm}$ and $\lambda = 400 \text{ nm}$:

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

The electron absorbs this photon, so its extra energy inside the metal is $3.1 \text{ eV}$.

During each collision, the electron loses $10\%$ of its extra energy to the metal, meaning it retains $90\%$ of its energy after each collision.

Energy remaining after two collisions is:

$E' = 3.1 \times (0.9)^2 = 3.1 \times 0.81 = 2.511 \text{ eV}$

To emerge from the metal, the electron must overcome the work function $\Phi = 2.2 \text{ eV}$.

The kinetic energy of the electron as it exits the metal is:

$K = E' - \Phi$

$K = 2.511 - 2.2 = 0.311 \text{ eV}$

Rounding to two decimal places, the kinetic energy is $0.31 \text{ eV}$.