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

The energy of the incident photon is given by:

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

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

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

When the electron absorbs the photon, it gains an extra energy of $3.1 \text{ eV}$. During each collision, the electron loses $10\%$ of its energy, meaning it retains $90\%$ of its energy.

After $n$ collisions, the remaining energy of the electron is:

$E_n = E \times (0.9)^n$

For the electron to emerge from the metal, its remaining energy must be greater than the work function of the metal ($\phi = 2.2 \text{ eV}$). Thus, we require:

$E_n > 2.2 \text{ eV}$

Calculating the remaining energy after successive collisions:
After $1$ collision: $E_1 = 3.1 \times 0.9 = 2.79 \text{ eV}$
After $2$ collisions: $E_2 = 2.79 \times 0.9 = 2.511 \text{ eV}$
After $3$ collisions: $E_3 = 2.511 \times 0.9 = 2.2599 \text{ eV}$
After $4$ collisions: $E_4 = 2.2599 \times 0.9 \approx 2.034 \text{ eV}$

Since $E_3 > 2.2 \text{ eV}$ and $E_4 < 2.2 \text{ eV}$, the electron can undergo a maximum of $3$ collisions and still have enough energy to emerge from the metal.

Answer: $3$