Given, Wavelength of ultraviolet ray, $\lambda = 3000 \mathring{\text{A}} = 3000 \times 10^{-10} \text{ m} = 3 \times 10^{-7} \text{ m}$ Work function, $\phi = 2.28 \text{ eV} = 2.28 \times 1.6 \times 10^{-19} \text{ J} = 3.648 \times 10^{-19} \text{ J}$ Planck's constant, $h = 6.62 \times 10^{-34} \text{ J s}$ Speed of light, $c = 3 \times 10^8 \text{ m/s}$ Mass of electron, $m = 9.1 \times 10^{-31} \text{ kg}$ According to Einstein's photoelectric equation, $$ \begin{aligned} \frac{hc}{\lambda} &= \phi + K_{\text{max}} \\ \frac{hc}{\lambda} &= \phi + \frac{1}{2}mv^2 \end{aligned} $$ Calculating the energy of the incident photon, $$ \begin{aligned} \frac{hc}{\lambda} &= \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} \\ &= 6.62 \times 10^{-19} \text{ J} \end{aligned} $$ Now, substituting the values into the photoelectric equation, $$ \begin{aligned} \frac{1}{2}mv^2 &= \frac{hc}{\lambda} - \phi \\ \frac{1}{2}mv^2 &= 6.62 \times 10^{-19} - 3.648 \times 10^{-19} \\ \frac{1}{2}mv^2 &= 2.972 \times 10^{-19} \text{ J} \end{aligned} $$ Solving for velocity $v$, $$ \begin{aligned} v^2 &= \frac{2 \times 2.972 \times 10^{-19}}{9.1 \times 10^{-31}} \\ v^2 &= 0.6531868 \times 10^{12} \\ v &= \sqrt{6.531868 \times 10^{11}} \\ v &\approx 8.082 \times 10^5 \text{ m/s} \end{aligned} $$ The velocity of the emitted electron is $8.08 \times 10^5 \text{ m/s}$.