A photocell has a cathode with a threshold wavelength of 480 nm…

Step 1. The work function. $\phi = \dfrac{1240\ \text{eV nm}}{480\ \text{nm}} = 2.58\ \text{eV}$.

Step 2. The photon energy. $E = \dfrac{1240\ \text{eV nm}}{365\ \text{nm}} = 3.40\ \text{eV}$.

Step 3. The largest kinetic energy. $K_{\max} = 3.40\ \text{eV} - 2.58\ \text{eV} = 0.81\ \text{eV}$.

Step 4. The stopping potential. $eV_0 = K_{\max}$, so $V_0 = 0.81$ V.

Step 5. The speed. $K_{\max} = 0.81 \times 1.6 \times 10^{-19}\ \text{J} = 1.30 \times 10^{-19}\ \text{J}$. We want $v$, so $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 1.30 \times 10^{-19}\ \text{J}}{9.1 \times 10^{-31}\ \text{kg}}} = 5.35 \times 10^{5}\ \text{m/s}$.