Translate the previous solution to the last problem you solved in English

Wavelength $\lambda = 620 \text{ nm} = 620 \times 10^{-9} \text{ m}$.  
Planck's constant $h = 6.63 \times 10^{-34} \text{ J s}$.  
Speed of light $c = 3.0 \times 10^8 \text{ m s}^{-1}$.

$$
E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{620 \times 10^{-9} \text{ m}}
$$
$$
E = \frac{1.989 \times 10^{-25}}{620 \times 10^{-9}} \text{ J}
$$
$$
E \approx 3.208 \times 10^{-19} \text{ J}
$$
$$
E \approx 3.2 \times 10^{-19} \text{ J}
$$
The energy of each photon is indeed $\boxed{3.2 \times 10^{-19} \text{ J}}$.

Power $P = I \times A$.  
Intensity $I = 5.0 \times 10^2 \text{ W m}^{-2}$.  
Area $A = 1.2 \times 10^{-3} \text{ m}^2$.
$$
P = (5.0 \times 10^2 \text{ W m}^{-2}) \times (1.2 \times 10^{-3} \text{ m}^2)
$$
$$
P = 0.60 \text{ W}
$$

The number of photons per second $N = \frac{P}{E}$.  
Energy per photon $E = 3.2 \times 10^{-19} \text{ J}$ (calculated in (i)).
$$
N = \frac{0.60 \text{ J s}^{-1}}{3.2 \times 10^{-19} \text{ J/photon}}
$$
$$
N \approx 1.875 \times 10^{18} \text{ photons s}^{-1}
$$
The number of photons falling on the metal surface each second is approximately $\boxed{1.9 \times 10^{18} \text{ photons s}^{-1}}$.

Work function $\phi = 2.4 \text{ eV}$.  
$1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$.
$$
\phi = 2.4 \times (1.602 \times 10^{-19} \text{ J})
$$
$$
\phi \approx 3.845 \times 10^{-19} \text{ J}
$$

Photon energy $E = 3.2 \times 10^{-19} \text{ J}$.  
Work function $\phi \approx 3.845 \times 10^{-19} \text{ J}$.  
Since $E < \phi$ ($3.2 \times 10^{-19} \text{ J} < 3.845 \times 10^{-19} \text{ J}$), the energy of the incident photons is insufficient to overcome the work function of the metal.  
Conclusion: $\boxed{\text{No electrons will be ejected}}$ from the metal surface.